summaryrefslogtreecommitdiff
path: root/doc/RecTutorial/RecTutorial.tex
blob: 9ee913d472b469986a2bb4e3614dd216ed412bef (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
1001
1002
1003
1004
1005
1006
1007
1008
1009
1010
1011
1012
1013
1014
1015
1016
1017
1018
1019
1020
1021
1022
1023
1024
1025
1026
1027
1028
1029
1030
1031
1032
1033
1034
1035
1036
1037
1038
1039
1040
1041
1042
1043
1044
1045
1046
1047
1048
1049
1050
1051
1052
1053
1054
1055
1056
1057
1058
1059
1060
1061
1062
1063
1064
1065
1066
1067
1068
1069
1070
1071
1072
1073
1074
1075
1076
1077
1078
1079
1080
1081
1082
1083
1084
1085
1086
1087
1088
1089
1090
1091
1092
1093
1094
1095
1096
1097
1098
1099
1100
1101
1102
1103
1104
1105
1106
1107
1108
1109
1110
1111
1112
1113
1114
1115
1116
1117
1118
1119
1120
1121
1122
1123
1124
1125
1126
1127
1128
1129
1130
1131
1132
1133
1134
1135
1136
1137
1138
1139
1140
1141
1142
1143
1144
1145
1146
1147
1148
1149
1150
1151
1152
1153
1154
1155
1156
1157
1158
1159
1160
1161
1162
1163
1164
1165
1166
1167
1168
1169
1170
1171
1172
1173
1174
1175
1176
1177
1178
1179
1180
1181
1182
1183
1184
1185
1186
1187
1188
1189
1190
1191
1192
1193
1194
1195
1196
1197
1198
1199
1200
1201
1202
1203
1204
1205
1206
1207
1208
1209
1210
1211
1212
1213
1214
1215
1216
1217
1218
1219
1220
1221
1222
1223
1224
1225
1226
1227
1228
1229
1230
1231
1232
1233
1234
1235
1236
1237
1238
1239
1240
1241
1242
1243
1244
1245
1246
1247
1248
1249
1250
1251
1252
1253
1254
1255
1256
1257
1258
1259
1260
1261
1262
1263
1264
1265
1266
1267
1268
1269
1270
1271
1272
1273
1274
1275
1276
1277
1278
1279
1280
1281
1282
1283
1284
1285
1286
1287
1288
1289
1290
1291
1292
1293
1294
1295
1296
1297
1298
1299
1300
1301
1302
1303
1304
1305
1306
1307
1308
1309
1310
1311
1312
1313
1314
1315
1316
1317
1318
1319
1320
1321
1322
1323
1324
1325
1326
1327
1328
1329
1330
1331
1332
1333
1334
1335
1336
1337
1338
1339
1340
1341
1342
1343
1344
1345
1346
1347
1348
1349
1350
1351
1352
1353
1354
1355
1356
1357
1358
1359
1360
1361
1362
1363
1364
1365
1366
1367
1368
1369
1370
1371
1372
1373
1374
1375
1376
1377
1378
1379
1380
1381
1382
1383
1384
1385
1386
1387
1388
1389
1390
1391
1392
1393
1394
1395
1396
1397
1398
1399
1400
1401
1402
1403
1404
1405
1406
1407
1408
1409
1410
1411
1412
1413
1414
1415
1416
1417
1418
1419
1420
1421
1422
1423
1424
1425
1426
1427
1428
1429
1430
1431
1432
1433
1434
1435
1436
1437
1438
1439
1440
1441
1442
1443
1444
1445
1446
1447
1448
1449
1450
1451
1452
1453
1454
1455
1456
1457
1458
1459
1460
1461
1462
1463
1464
1465
1466
1467
1468
1469
1470
1471
1472
1473
1474
1475
1476
1477
1478
1479
1480
1481
1482
1483
1484
1485
1486
1487
1488
1489
1490
1491
1492
1493
1494
1495
1496
1497
1498
1499
1500
1501
1502
1503
1504
1505
1506
1507
1508
1509
1510
1511
1512
1513
1514
1515
1516
1517
1518
1519
1520
1521
1522
1523
1524
1525
1526
1527
1528
1529
1530
1531
1532
1533
1534
1535
1536
1537
1538
1539
1540
1541
1542
1543
1544
1545
1546
1547
1548
1549
1550
1551
1552
1553
1554
1555
1556
1557
1558
1559
1560
1561
1562
1563
1564
1565
1566
1567
1568
1569
1570
1571
1572
1573
1574
1575
1576
1577
1578
1579
1580
1581
1582
1583
1584
1585
1586
1587
1588
1589
1590
1591
1592
1593
1594
1595
1596
1597
1598
1599
1600
1601
1602
1603
1604
1605
1606
1607
1608
1609
1610
1611
1612
1613
1614
1615
1616
1617
1618
1619
1620
1621
1622
1623
1624
1625
1626
1627
1628
1629
1630
1631
1632
1633
1634
1635
1636
1637
1638
1639
1640
1641
1642
1643
1644
1645
1646
1647
1648
1649
1650
1651
1652
1653
1654
1655
1656
1657
1658
1659
1660
1661
1662
1663
1664
1665
1666
1667
1668
1669
1670
1671
1672
1673
1674
1675
1676
1677
1678
1679
1680
1681
1682
1683
1684
1685
1686
1687
1688
1689
1690
1691
1692
1693
1694
1695
1696
1697
1698
1699
1700
1701
1702
1703
1704
1705
1706
1707
1708
1709
1710
1711
1712
1713
1714
1715
1716
1717
1718
1719
1720
1721
1722
1723
1724
1725
1726
1727
1728
1729
1730
1731
1732
1733
1734
1735
1736
1737
1738
1739
1740
1741
1742
1743
1744
1745
1746
1747
1748
1749
1750
1751
1752
1753
1754
1755
1756
1757
1758
1759
1760
1761
1762
1763
1764
1765
1766
1767
1768
1769
1770
1771
1772
1773
1774
1775
1776
1777
1778
1779
1780
1781
1782
1783
1784
1785
1786
1787
1788
1789
1790
1791
1792
1793
1794
1795
1796
1797
1798
1799
1800
1801
1802
1803
1804
1805
1806
1807
1808
1809
1810
1811
1812
1813
1814
1815
1816
1817
1818
1819
1820
1821
1822
1823
1824
1825
1826
1827
1828
1829
1830
1831
1832
1833
1834
1835
1836
1837
1838
1839
1840
1841
1842
1843
1844
1845
1846
1847
1848
1849
1850
1851
1852
1853
1854
1855
1856
1857
1858
1859
1860
1861
1862
1863
1864
1865
1866
1867
1868
1869
1870
1871
1872
1873
1874
1875
1876
1877
1878
1879
1880
1881
1882
1883
1884
1885
1886
1887
1888
1889
1890
1891
1892
1893
1894
1895
1896
1897
1898
1899
1900
1901
1902
1903
1904
1905
1906
1907
1908
1909
1910
1911
1912
1913
1914
1915
1916
1917
1918
1919
1920
1921
1922
1923
1924
1925
1926
1927
1928
1929
1930
1931
1932
1933
1934
1935
1936
1937
1938
1939
1940
1941
1942
1943
1944
1945
1946
1947
1948
1949
1950
1951
1952
1953
1954
1955
1956
1957
1958
1959
1960
1961
1962
1963
1964
1965
1966
1967
1968
1969
1970
1971
1972
1973
1974
1975
1976
1977
1978
1979
1980
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
2010
2011
2012
2013
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
2024
2025
2026
2027
2028
2029
2030
2031
2032
2033
2034
2035
2036
2037
2038
2039
2040
2041
2042
2043
2044
2045
2046
2047
2048
2049
2050
2051
2052
2053
2054
2055
2056
2057
2058
2059
2060
2061
2062
2063
2064
2065
2066
2067
2068
2069
2070
2071
2072
2073
2074
2075
2076
2077
2078
2079
2080
2081
2082
2083
2084
2085
2086
2087
2088
2089
2090
2091
2092
2093
2094
2095
2096
2097
2098
2099
2100
2101
2102
2103
2104
2105
2106
2107
2108
2109
2110
2111
2112
2113
2114
2115
2116
2117
2118
2119
2120
2121
2122
2123
2124
2125
2126
2127
2128
2129
2130
2131
2132
2133
2134
2135
2136
2137
2138
2139
2140
2141
2142
2143
2144
2145
2146
2147
2148
2149
2150
2151
2152
2153
2154
2155
2156
2157
2158
2159
2160
2161
2162
2163
2164
2165
2166
2167
2168
2169
2170
2171
2172
2173
2174
2175
2176
2177
2178
2179
2180
2181
2182
2183
2184
2185
2186
2187
2188
2189
2190
2191
2192
2193
2194
2195
2196
2197
2198
2199
2200
2201
2202
2203
2204
2205
2206
2207
2208
2209
2210
2211
2212
2213
2214
2215
2216
2217
2218
2219
2220
2221
2222
2223
2224
2225
2226
2227
2228
2229
2230
2231
2232
2233
2234
2235
2236
2237
2238
2239
2240
2241
2242
2243
2244
2245
2246
2247
2248
2249
2250
2251
2252
2253
2254
2255
2256
2257
2258
2259
2260
2261
2262
2263
2264
2265
2266
2267
2268
2269
2270
2271
2272
2273
2274
2275
2276
2277
2278
2279
2280
2281
2282
2283
2284
2285
2286
2287
2288
2289
2290
2291
2292
2293
2294
2295
2296
2297
2298
2299
2300
2301
2302
2303
2304
2305
2306
2307
2308
2309
2310
2311
2312
2313
2314
2315
2316
2317
2318
2319
2320
2321
2322
2323
2324
2325
2326
2327
2328
2329
2330
2331
2332
2333
2334
2335
2336
2337
2338
2339
2340
2341
2342
2343
2344
2345
2346
2347
2348
2349
2350
2351
2352
2353
2354
2355
2356
2357
2358
2359
2360
2361
2362
2363
2364
2365
2366
2367
2368
2369
2370
2371
2372
2373
2374
2375
2376
2377
2378
2379
2380
2381
2382
2383
2384
2385
2386
2387
2388
2389
2390
2391
2392
2393
2394
2395
2396
2397
2398
2399
2400
2401
2402
2403
2404
2405
2406
2407
2408
2409
2410
2411
2412
2413
2414
2415
2416
2417
2418
2419
2420
2421
2422
2423
2424
2425
2426
2427
2428
2429
2430
2431
2432
2433
2434
2435
2436
2437
2438
2439
2440
2441
2442
2443
2444
2445
2446
2447
2448
2449
2450
2451
2452
2453
2454
2455
2456
2457
2458
2459
2460
2461
2462
2463
2464
2465
2466
2467
2468
2469
2470
2471
2472
2473
2474
2475
2476
2477
2478
2479
2480
2481
2482
2483
2484
2485
2486
2487
2488
2489
2490
2491
2492
2493
2494
2495
2496
2497
2498
2499
2500
2501
2502
2503
2504
2505
2506
2507
2508
2509
2510
2511
2512
2513
2514
2515
2516
2517
2518
2519
2520
2521
2522
2523
2524
2525
2526
2527
2528
2529
2530
2531
2532
2533
2534
2535
2536
2537
2538
2539
2540
2541
2542
2543
2544
2545
2546
2547
2548
2549
2550
2551
2552
2553
2554
2555
2556
2557
2558
2559
2560
2561
2562
2563
2564
2565
2566
2567
2568
2569
2570
2571
2572
2573
2574
2575
2576
2577
2578
2579
2580
2581
2582
2583
2584
2585
2586
2587
2588
2589
2590
2591
2592
2593
2594
2595
2596
2597
2598
2599
2600
2601
2602
2603
2604
2605
2606
2607
2608
2609
2610
2611
2612
2613
2614
2615
2616
2617
2618
2619
2620
2621
2622
2623
2624
2625
2626
2627
2628
2629
2630
2631
2632
2633
2634
2635
2636
2637
2638
2639
2640
2641
2642
2643
2644
2645
2646
2647
2648
2649
2650
2651
2652
2653
2654
2655
2656
2657
2658
2659
2660
2661
2662
2663
2664
2665
2666
2667
2668
2669
2670
2671
2672
2673
2674
2675
2676
2677
2678
2679
2680
2681
2682
2683
2684
2685
2686
2687
2688
2689
2690
2691
2692
2693
2694
2695
2696
2697
2698
2699
2700
2701
2702
2703
2704
2705
2706
2707
2708
2709
2710
2711
2712
2713
2714
2715
2716
2717
2718
2719
2720
2721
2722
2723
2724
2725
2726
2727
2728
2729
2730
2731
2732
2733
2734
2735
2736
2737
2738
2739
2740
2741
2742
2743
2744
2745
2746
2747
2748
2749
2750
2751
2752
2753
2754
2755
2756
2757
2758
2759
2760
2761
2762
2763
2764
2765
2766
2767
2768
2769
2770
2771
2772
2773
2774
2775
2776
2777
2778
2779
2780
2781
2782
2783
2784
2785
2786
2787
2788
2789
2790
2791
2792
2793
2794
2795
2796
2797
2798
2799
2800
2801
2802
2803
2804
2805
2806
2807
2808
2809
2810
2811
2812
2813
2814
2815
2816
2817
2818
2819
2820
2821
2822
2823
2824
2825
2826
2827
2828
2829
2830
2831
2832
2833
2834
2835
2836
2837
2838
2839
2840
2841
2842
2843
2844
2845
2846
2847
2848
2849
2850
2851
2852
2853
2854
2855
2856
2857
2858
2859
2860
2861
2862
2863
2864
2865
2866
2867
2868
2869
2870
2871
2872
2873
2874
2875
2876
2877
2878
2879
2880
2881
2882
2883
2884
2885
2886
2887
2888
2889
2890
2891
2892
2893
2894
2895
2896
2897
2898
2899
2900
2901
2902
2903
2904
2905
2906
2907
2908
2909
2910
2911
2912
2913
2914
2915
2916
2917
2918
2919
2920
2921
2922
2923
2924
2925
2926
2927
2928
2929
2930
2931
2932
2933
2934
2935
2936
2937
2938
2939
2940
2941
2942
2943
2944
2945
2946
2947
2948
2949
2950
2951
2952
2953
2954
2955
2956
2957
2958
2959
2960
2961
2962
2963
2964
2965
2966
2967
2968
2969
2970
2971
2972
2973
2974
2975
2976
2977
2978
2979
2980
2981
2982
2983
2984
2985
2986
2987
2988
2989
2990
2991
2992
2993
2994
2995
2996
2997
2998
2999
3000
3001
3002
3003
3004
3005
3006
3007
3008
3009
3010
3011
3012
3013
3014
3015
3016
3017
3018
3019
3020
3021
3022
3023
3024
3025
3026
3027
3028
3029
3030
3031
3032
3033
3034
3035
3036
3037
3038
3039
3040
3041
3042
3043
3044
3045
3046
3047
3048
3049
3050
3051
3052
3053
3054
3055
3056
3057
3058
3059
3060
3061
3062
3063
3064
3065
3066
3067
3068
3069
3070
3071
3072
3073
3074
3075
3076
3077
3078
3079
3080
3081
3082
3083
3084
3085
3086
3087
3088
3089
3090
3091
3092
3093
3094
3095
3096
3097
3098
3099
3100
3101
3102
3103
3104
3105
3106
3107
3108
3109
3110
3111
3112
3113
3114
3115
3116
3117
3118
3119
3120
3121
3122
3123
3124
3125
3126
3127
3128
3129
3130
3131
3132
3133
3134
3135
3136
3137
3138
3139
3140
3141
3142
3143
3144
3145
3146
3147
3148
3149
3150
3151
3152
3153
3154
3155
3156
3157
3158
3159
3160
3161
3162
3163
3164
3165
3166
3167
3168
3169
3170
3171
3172
3173
3174
3175
3176
3177
3178
3179
3180
3181
3182
3183
3184
3185
3186
3187
3188
3189
3190
3191
3192
3193
3194
3195
3196
3197
3198
3199
3200
3201
3202
3203
3204
3205
3206
3207
3208
3209
3210
3211
3212
3213
3214
3215
3216
3217
3218
3219
3220
3221
3222
3223
3224
3225
3226
3227
3228
3229
3230
3231
3232
3233
3234
3235
3236
3237
3238
3239
3240
3241
3242
3243
3244
3245
3246
3247
3248
3249
3250
3251
3252
3253
3254
3255
3256
3257
3258
3259
3260
3261
3262
3263
3264
3265
3266
3267
3268
3269
3270
3271
3272
3273
3274
3275
3276
3277
3278
3279
3280
3281
3282
3283
3284
3285
3286
3287
3288
3289
3290
3291
3292
3293
3294
3295
3296
3297
3298
3299
3300
3301
3302
3303
3304
3305
3306
3307
3308
3309
3310
3311
3312
3313
3314
3315
3316
3317
3318
3319
3320
3321
3322
3323
3324
3325
3326
3327
3328
3329
3330
3331
3332
3333
3334
3335
3336
3337
3338
3339
3340
3341
3342
3343
3344
3345
3346
3347
3348
3349
3350
3351
3352
3353
3354
3355
3356
3357
3358
3359
3360
3361
3362
3363
3364
3365
3366
3367
3368
3369
3370
3371
3372
3373
3374
3375
3376
3377
3378
3379
3380
3381
3382
3383
3384
3385
3386
3387
3388
3389
3390
3391
3392
3393
3394
3395
3396
3397
3398
3399
3400
3401
3402
3403
3404
3405
3406
3407
3408
3409
3410
3411
3412
3413
3414
3415
3416
3417
3418
3419
3420
3421
3422
3423
3424
3425
3426
3427
3428
3429
3430
3431
3432
3433
3434
3435
3436
3437
3438
3439
3440
3441
3442
3443
3444
3445
3446
3447
3448
3449
3450
3451
3452
3453
3454
3455
3456
3457
3458
3459
3460
3461
3462
3463
3464
3465
3466
3467
3468
3469
3470
3471
3472
3473
3474
3475
3476
3477
3478
3479
3480
3481
3482
3483
3484
3485
3486
3487
3488
3489
3490
3491
3492
3493
3494
3495
3496
3497
3498
3499
3500
3501
3502
3503
3504
3505
3506
3507
3508
3509
3510
3511
3512
3513
3514
3515
3516
3517
3518
3519
3520
3521
3522
3523
3524
3525
3526
3527
3528
3529
3530
3531
3532
3533
3534
3535
3536
3537
3538
3539
3540
3541
3542
3543
3544
3545
3546
3547
3548
3549
3550
3551
3552
3553
3554
3555
3556
3557
3558
3559
3560
3561
3562
3563
3564
3565
3566
3567
3568
3569
3570
3571
3572
3573
3574
3575
3576
3577
3578
3579
3580
3581
3582
3583
3584
3585
3586
3587
3588
3589
3590
3591
3592
3593
3594
3595
3596
3597
3598
3599
3600
3601
3602
3603
3604
3605
3606
\documentclass[11pt]{article}
\title{A Tutorial on [Co-]Inductive Types in Coq}
\author{Eduardo Gim\'enez\thanks{Eduardo.Gimenez@inria.fr},
Pierre Cast\'eran\thanks{Pierre.Casteran@labri.fr}}
\date{May 1998 --- \today}

\usepackage{multirow}
\usepackage{aeguill}
%\externaldocument{RefMan-gal.v}
%\externaldocument{RefMan-ext.v}
%\externaldocument{RefMan-tac.v}
%\externaldocument{RefMan-oth}
%\externaldocument{RefMan-tus.v}
%\externaldocument{RefMan-syn.v}
%\externaldocument{Extraction.v}
\input{recmacros}
\input{coqartmacros}
\newcommand{\refmancite}[1]{{}}
%\newcommand{\refmancite}[1]{\cite{coqrefman}}
%\newcommand{\refmancite}[1]{\cite[#1] {]{coqrefman}}

\usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{makeidx}
%\usepackage{multind}
\usepackage{alltt}   
\usepackage{verbatim}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{theorem}
\usepackage[dvips]{epsfig}
\usepackage{epic}
\usepackage{eepic}
\usepackage{ecltree}
\usepackage{moreverb} 
\usepackage{color}
\usepackage{pifont}
\usepackage{xr}
\usepackage{url}

\usepackage{alltt}
\renewcommand{\familydefault}{ptm}
\renewcommand{\seriesdefault}{m}
\renewcommand{\shapedefault}{n}
\newtheorem{exercise}{Exercise}[section]	
\makeindex
\begin{document}
\maketitle

\begin{abstract}
This document\footnote{The first versions of this document were entirely written by Eduardo Gimenez.
Pierre Cast\'eran wrote the 2004 revision.} is an introduction to the definition and
use of inductive and co-inductive  types in the {\coq} proof environment. It explains how types like natural numbers and infinite streams are defined
in {\coq}, and the kind of proof techniques that can be used to reason
about them (case analysis, induction, inversion of predicates,
co-induction, etc).  Each technique is illustrated through an
executable and self-contained {\coq} script.  
\end{abstract}
%\RRkeyword{Proof environments, recursive types.}
%\makeRT

\addtocontents{toc}{\protect \thispagestyle{empty}}
\pagenumbering{arabic}

\cleardoublepage
\tableofcontents
\clearpage

\section{About this document}

This document is an introduction to the definition and use of
inductive and co-inductive  types in the {\coq} proof environment.  It was born from the
notes written for the course about the version V5.10 of {\coq}, given
by Eduardo Gimenez  at
the Ecole Normale Sup\'erieure de Lyon in March 1996. This article is
a revised and improved version of these notes for the version V8.0 of
the system.


We assume that the reader has some familiarity with the
proofs-as-programs paradigm of Logic \cite{Coquand:metamathematical} and the generalities
of the {\coq} system \cite{coqrefman}.  You would take a greater advantage of
this document if you first read the general tutorial about {\coq} and
{\coq}'s FAQ, both available on \cite{coqsite}.
A text book \cite{coqart},  accompanied with a lot of
examples and exercises \cite{Booksite}, presents a detailed description
of the {\coq} system and its underlying
formalism: the Calculus of Inductive Construction.
Finally, the complete description of {\coq}  is given in the reference manual
\cite{coqrefman}. Most of the tactics and commands we describe have
several options, which we do not present exhaustively. 
If some script herein uses a non described feature, please refer to
the Reference Manual.


If you are familiar with other proof environments
based on type theory and the LCF style ---like PVS, LEGO, Isabelle,
etc--- then you will find not difficulty to guess the unexplained
details.

The better way to read this document is to start up the {\coq} system,
type by yourself the examples and exercises, and observe the
behavior of the system. All the examples proposed in this tutorial
can be downloaded from the same site as the present document. 


The tutorial is organised as follows. The next section describes how
inductive types are defined in {\coq}, and introduces some useful ones,
like natural numbers, the empty type, the propositional equality type,
and the logical connectives. Section \ref{CaseAnalysis} explains
definitions by pattern-matching and their connection with the
principle of case analysis. This principle is the most basic
elimination rule associated with inductive or co-inductive types
 and follows a
general scheme that we illustrate for some of the types introduced in
Section \ref{Introduction}.  Section \ref{CaseTechniques} illustrates
the pragmatics of this principle, showing different proof techniques
based on it. Section \ref{StructuralInduction} introduces definitions
by structural recursion and proofs by induction. 
Section~\ref{CaseStudy} presents some elaborate techniques
about dependent case analysis. Finally, Section
\ref{CoInduction} is a brief introduction to co-inductive types
--i.e., types containing infinite objects-- and the principle of
co-induction.

Thanks to  Bruno Barras, Yves Bertot, Hugo Herbelin, Jean-Fran\c{c}ois Monin
and Michel L\'evy for their help.

\subsection*{Lexical conventions}
The \texttt{typewriter} font is used to represent text
input by the user, while the \textit{italic} font is used to represent
the text output by the system as answers.  


Moreover, the mathematical symbols \coqle{}, \coqdiff, \(\exists\),
\(\forall\), \arrow{}, $\rightarrow{}$ \coqor{}, \coqand{}, and \funarrow{} 
stand for the character strings \citecoq{<=}, \citecoq{<>},
\citecoq{exists}, \citecoq{forall}, \citecoq{->}, \citecoq{<-},
\texttt{\char'134/}, \texttt{/\char'134}, and \citecoq{=>},
respectively.  For instance, the \coq{} statement
%V8 A prendre
% inclusion numero 1
% traduction numero 1
\begin{alltt}
\hide{Open Scope nat_scope. Check (}forall A:Set,(exists x : A, forall (y:A), x <> y) -> 2 = 3\hide{).}
\end{alltt}
is written as follows in this tutorial:
%V8 A prendre
% inclusion numero 2
% traduction numero 2
\begin{alltt}
\hide{Check (}{\prodsym}A:Set,(\exsym{}x:A, {\prodsym}y:A, x {\coqdiff} y) \arrow{} 2 = 3\hide{).}
\end{alltt}

When a fragment of \coq{} input text appears in the middle of
regular text, we often place this fragment between double quotes
``\dots.''  These double quotes do not belong to the \coq{} syntax.

Finally, any
string enclosed between \texttt{(*} and \texttt{*)} is a comment and
is ignored by the \coq{} system.

\section{Introducing Inductive Types} 
\label{Introduction}

Inductive types are types closed with respect to their introduction
rules. These rules explain the most basic or \textsl{canonical} ways
of constructing an element of the type.  In this sense, they
characterize the recursive type. Different rules must be considered as
introducing different objects. In order to fix ideas, let us introduce
in {\coq} the most well-known example of a recursive type: the type of
natural numbers.

%V8 A prendre
\begin{alltt}
Inductive nat : Set := 
 | O : nat 
 | S : nat\arrow{}nat.
\end{alltt}

The definition of a recursive type has two main parts. First, we
establish what kind of recursive type we will characterize (a set, in
this case). Second, we present the introduction rules that define the
type ({\Z} and {\SUCC}), also called its {\sl constructors}. The constructors
{\Z} and {\SUCC} determine all the elements of this type. In other
words, if $n\mbox{:}\nat$, then $n$ must have been introduced either
by the rule {\Z} or by an application of the rule {\SUCC} to a
previously constructed natural number. In this sense, we can say
that {\nat} is \emph{closed}. On the contrary, the type
$\Set$ is an {\it open} type, since we do not know {\it a priori} all
the possible ways of introducing an object of type \texttt{Set}.

After entering this command, the constants {\nat}, {\Z} and {\SUCC} are
available in the current context. We can see their types using the
\texttt{Check} command \refmancite{Section \ref{Check}}:

%V8 A prendre 
\begin{alltt}
Check nat.
\it{}nat : Set
\tt{}Check O.
\it{}O : nat
\tt{}Check S.
\it{}S : nat {\arrow} nat
\end{alltt}

Moreover, {\coq} adds to the context three constants named
 $\natind$, $\natrec$ and $\natrect$, which
 correspond to different principles of structural induction on
natural numbers that {\coq} infers automatically from the definition.  We
will come back to them in Section \ref{StructuralInduction}.


In fact, the type of natural numbers as well as several useful
theorems about them are already defined in the basic library of {\coq},
so there is no need to introduce them.  Therefore, let us throw away
our (re)definition of {\nat}, using the command \texttt{Reset}.

%V8 A prendre
\begin{alltt}
Reset nat.
Print nat.
\it{}Inductive nat : Set :=  O : nat | S : nat \arrow{} nat
For S: Argument scope is [nat_scope]
\end{alltt}

Notice that \coq{}'s \emph{interpretation scope} for natural numbers
(called \texttt{nat\_scope}) 
allows us to read and write natural numbers in decimal form (see \cite{coqrefman}). For instance, the constructor \texttt{O} can be read or written
as the digit $0$, and the term ``~\texttt{S (S (S O))}~'' as $3$.

%V8 A prendre
\begin{alltt}
Check O.
\it 0 : nat.
\tt
Check (S (S (S O))).
\it 3 : nat
\end{alltt}

Let us now take a look to some other
recursive types contained in the standard library of {\coq}.

\subsection{Lists}
Lists are defined in library \citecoq{List}:

\begin{alltt}
Require Import List.
Print list.
\it
Inductive list (A : Set) : Set :=
    nil : list A | cons : A {\arrow} list A {\arrow} list A
For nil: Argument A is implicit
For cons: Argument A is implicit
For list: Argument scope is [type_scope]
For nil: Argument scope is [type_scope]
For cons: Argument scopes are [type_scope _ _]
\end{alltt}

In this definition, \citecoq{A} is a \emph{general parameter}, global
to both constructors.
This kind of definition allows us to build a whole family of
inductive types, indexed over the sort \citecoq{Set}.
This can be observed if we consider the type of identifiers
\citecoq{list}, \citecoq{cons} and \citecoq{nil}.
Notice the notation \citecoq{(A := \dots)} which must be used 
when {\coq}'s type inference algorithm cannot infer the implicit
parameter \citecoq{A}.
\begin{alltt}
Check list.
\it list
     : Set {\arrow} Set

\tt Check (nil (A:=nat)).
\it nil
     : list nat

\tt Check (nil (A:= nat {\arrow} nat)).
\it nil
     : list (nat {\arrow} nat)

\tt Check (fun A: Set {\funarrow} (cons (A:=A))).
\it fun A : Set {\funarrow} cons (A:=A)
     : {\prodsym} A : Set, A {\arrow} list A {\arrow} list A

\tt Check (cons 3 (cons 2 nil)).
\it 3 :: 2 :: nil
     : list nat
\end{alltt}

\subsection{Vectors.}
\label{vectors}

Like \texttt{list}, \citecoq{vector} is a polymorphic type:
if $A$ is a set, and $n$ a natural number, ``~\citecoq{vector $A$ $n$}~''
is the type of vectors of elements of $A$ and size $n$.


\begin{alltt}
Require Import  Bvector.

Print vector.
\it
Inductive vector (A : Set) : nat {\arrow} Set :=
    Vnil : vector A 0
  | Vcons : A {\arrow} {\prodsym} n : nat, vector A n {\arrow} vector A (S n)
For vector: Argument scopes are [type_scope nat_scope]
For Vnil: Argument scope is [type_scope]
For Vcons: Argument scopes are [type_scope _ nat_scope _]
\end{alltt}


Remark the difference between the two parameters $A$ and $n$:
The first one is a \textsl{general parameter}, global to all the
introduction rules,while the second one is an \textsl{index}, which is
instantiated differently in the introduction rules.
Such types parameterized  by regular
values are called \emph{dependent types}.

\begin{alltt}
Check (Vnil nat).
\it Vnil nat
     : vector nat 0

\tt Check (fun (A:Set)(a:A){\funarrow} Vcons _ a _ (Vnil _)).
\it fun (A : Set) (a : A) {\funarrow} Vcons A a 0 (Vnil A)
     : {\prodsym} A : Set, A {\arrow} vector A 1


\tt Check (Vcons _ 5 _ (Vcons _ 3 _ (Vnil _))).
\it Vcons nat 5 1 (Vcons nat 3 0 (Vnil nat))
     : vector nat 2
\end{alltt}

\subsection{The contradictory proposition.}
Another example of an inductive type is the contradictory proposition.
This type inhabits the universe of propositions, and has no element
at all.
%V8 A prendre
\begin{alltt}
Print False.
\it{} Inductive False : Prop :=
\end{alltt}

\noindent Notice that no constructor is given in this definition.

\subsection{The tautological proposition.}
Similarly, the
tautological proposition {\True} is defined as an inductive type
with only one element {\I}:

%V8 A prendre
\begin{alltt}
Print True.
\it{}Inductive True : Prop :=  I : True
\end{alltt}

\subsection{Relations as inductive types.}
Some relations can also be introduced in a smart way as an inductive family
of propositions. Let us take as example the order $n \leq m$ on natural
numbers, called \citecoq{le} in {\coq}.
 This relation is introduced through
the following definition, quoted from the standard library\footnote{In the interpretation scope
for Peano arithmetic:
\citecoq{nat\_scope},  ``~\citecoq{n <= m}~'' is equivalent to 
``~\citecoq{le n m}~'' .}:




%V8 A prendre
\begin{alltt}
Print le. \it
Inductive le (n:nat) : nat\arrow{}Prop := 
|  le_n: n {\coqle} n 
|  le_S: {\prodsym} m, n {\coqle} m \arrow{} n {\coqle} S m.
\end{alltt}

Notice that in this definition $n$ is a general parameter,
while the second argument of \citecoq{le} is an index (see section
~\ref{vectors}).
 This definition
introduces the binary relation $n {\leq} m$ as the family of unary predicates
``\textsl{to be greater or equal than a given $n$}'', parameterized by $n$.

The introduction rules of this type can be seen as a sort of Prolog
rules for proving that a given integer $n$ is less or equal than another one.
In fact, an object of type $n{\leq} m$ is nothing but a proof 
built up using the constructors \textsl{le\_n} and
\textsl{le\_S} of this type.  As an example, let us construct
a proof that zero is less or equal than three using {\coq}'s interactive
proof mode.
Such an object can be obtained applying three times the second
introduction rule of \citecoq{le}, to a proof that zero is less or equal
than itself,
which is provided by the first constructor of \citecoq{le}:

%V8 A prendre
\begin{alltt}
Theorem zero_leq_three: 0 {\coqle} 3.
Proof.
\it{} 1 subgoal

============================
 0 {\coqle} 3

\tt{}Proof.
 constructor 2. 

\it{} 1 subgoal
============================
  0 {\coqle} 2

\tt{} constructor 2.  
\it{} 1 subgoal
============================
  0 {\coqle} 1

\tt{} constructor 2
\it{} 1 subgoal
============================
  0 {\coqle} 0

\tt{} constructor 1.

\it{}Proof completed
\tt{}Qed.
\end{alltt}

\noindent When
the current goal is an inductive type, the tactic 
``~\citecoq{constructor $i$}~'' \refmancite{Section \ref{constructor}} applies the $i$-th constructor in the
definition of the type. We can take a look at the proof constructed
using the command \texttt{Print}:

%V8 A prendre
\begin{alltt}
Print Print zero_leq_three.
\it{}zero_leq_three = 
zero_leq_three = le_S 0 2 (le_S 0 1 (le_S 0 0 (le_n 0)))
     : 0 {\coqle} 3
\end{alltt}

When the parameter $i$ is not supplied, the tactic \texttt{constructor}
tries to apply ``~\texttt{constructor $1$}~'', ``~\texttt{constructor $2$}~'',\dots,
``~\texttt{constructor $n$}~'' where $n$ is the number of constructors
of the inductive type (2 in our example) of the conclusion of the goal.
Our little proof can thus be obtained iterating the tactic
\texttt{constructor} until it fails:

%V8 A prendre
\begin{alltt}
Lemma zero_leq_three': 0 {\coqle} 3.
 repeat constructor.
Qed.
\end{alltt}

Notice that the strict order on \texttt{nat}, called \citecoq{lt}
is not inductively defined:

\begin{alltt}
Print lt.
\it
lt = fun n m : nat {\funarrow} S n {\coqle} m
     : nat {\arrow} nat {\arrow} Prop
\tt
Lemma zero_lt_three : 0 < 3.
Proof.
 unfold lt.
\it
====================
 1 {\coqle} 3
\tt
 repeat constructor. 
Qed.
\end{alltt}



\subsection{The propositional equality type.} \label{equality}
In {\coq}, the propositional equality between two inhabitants $a$ and
$b$ of
the same type $A$ ,
noted $a=b$, is introduced as a family of recursive predicates
``~\textsl{to be equal to $a$}~'',  parameterised by both $a$ and its type
$A$. This family of types has only one introduction rule, which
corresponds to reflexivity.
Notice that the syntax ``\citecoq{$a$ = $b$}~'' is an abbreviation  
for ``\citecoq{eq $a$  $b$}~'', and that the parameter $A$ is \emph{implicit},
as it can be infered from $a$.
%V8 A prendre
\begin{alltt}
Print eq.
\it{} Inductive eq (A : Type) (x : A) : A \arrow{} Prop :=  
    refl_equal : x = x
For eq: Argument A is implicit
For refl_equal: Argument A is implicit
For eq: Argument scopes are [type_scope _ _]
For refl_equal: Argument scopes are [type_scope _]
\end{alltt}

Notice also that  the first parameter $A$ of \texttt{eq} has type
\texttt{Type}. The type system of {\coq} allows us to consider equality between 
various kinds of terms: elements of a set, proofs, propositions,
types, and so on.
Look at \cite{coqrefman, coqart} to get more details on {\coq}'s type
system, as well as implicit arguments and argument scopes.


\begin{alltt}
Lemma eq_3_3 : 2 + 1 = 3.
Proof.
 reflexivity.
Qed.

Lemma eq_proof_proof : refl_equal (2*6) = refl_equal (3*4).
Proof.
 reflexivity.
Qed.

Print eq_proof_proof.
\it eq_proof_proof = 
refl_equal (refl_equal (3 * 4))
     : refl_equal (2 * 6) = refl_equal (3 * 4)
\tt

Lemma eq_lt_le : ( 2 < 4) = (3 {\coqle} 4).
Proof.
 reflexivity.
Qed.

Lemma eq_nat_nat : nat = nat.
Proof.
 reflexivity.
Qed.

Lemma eq_Set_Set : Set = Set.
Proof.
 reflexivity.
Qed.
\end{alltt}

\subsection{Logical connectives.} \label{LogicalConnectives}
The conjunction and disjunction of two propositions are also examples
of recursive types:

\begin{alltt}
Inductive or (A B : Prop) : Prop :=
    or_introl : A \arrow{} A {\coqor} B | or_intror : B \arrow{} A {\coqor} B

Inductive and (A B : Prop) : Prop :=  
    conj : A \arrow{} B \arrow{} A {\coqand} B

\end{alltt}

The propositions $A$ and $B$ are general parameters of these
connectives. Choosing different universes for 
$A$ and $B$ and for the inductive type itself gives rise to different
type constructors. For example, the type \textsl{sumbool} is a
disjunction but with computational contents.

\begin{alltt}
Inductive sumbool (A B : Prop) : Set :=
    left : A \arrow{} \{A\} + \{B\} | right : B \arrow{} \{A\} + \{B\}
\end{alltt}



This type --noted \texttt{\{$A$\}+\{$B$\}} in {\coq}-- can be used in {\coq}
programs as a sort of boolean type, to check whether it is $A$ or $B$
that is true.  The values ``~\citecoq{left $p$}~'' and
``~\citecoq{right $q$}~'' replace the boolean values \textsl{true} and
\textsl{false}, respectively. The advantage of this type over
\textsl{bool} is that it makes available the proofs $p$ of $A$ or $q$
of $B$, which could be necessary to construct a verification proof
about the program.
For instance, let us consider the certified program \citecoq{le\_lt\_dec}
of the Standard Library.

\begin{alltt}
Require Import Compare_dec.
Check le_lt_dec.
\it
le_lt_dec
     : {\prodsym} n m : nat, \{n {\coqle} m\} + \{m < n\}

\end{alltt}

We use \citecoq{le\_lt\_dec} to build a function for computing
the max of two natural numbers:

\begin{alltt}
Definition max (n p :nat) := match le_lt_dec n p with 
                             | left _ {\funarrow} p
                             | right _ {\funarrow} n
                             end.
\end{alltt}

In the following proof, the case analysis on the term
``~\citecoq{le\_lt\_dec n p}~'' gives us an access to proofs
of $n\leq p$ in the first case, $p<n$ in the other.

\begin{alltt}
Theorem le_max : {\prodsym} n p, n {\coqle} p {\arrow} max n p = p.
Proof.
 intros n p ; unfold max ; case (le_lt_dec n p); simpl.
\it
2 subgoals
  
  n : nat
  p : nat
  ============================
   n {\coqle} p {\arrow} n {\coqle} p {\arrow} p = p

subgoal 2 is:
 p < n {\arrow} n {\coqle} p {\arrow} n = p
\tt
 trivial.
 intros; absurd (p < p); eauto with arith.
Qed.
\end{alltt}


 Once the program verified, the proofs are
erased by the extraction procedure:

\begin{alltt}
Extraction max.
\it
(** val max : nat {\arrow} nat {\arrow} nat **)

let max n p =
  match le_lt_dec n p with
    | Left {\arrow} p
    | Right {\arrow} n
\end{alltt}

Another example of use of \citecoq{sumbool} is given  in Section
\ref{WellFoundedRecursion}.

\subsection{The existential quantifier.}\label{ex-def}
The existential quantifier is yet another example of a logical
connective introduced as an inductive type.

\begin{alltt}
Inductive ex (A : Type) (P : A \arrow{} Prop) : Prop :=
    ex_intro : {\prodsym} x : A, P x \arrow{} ex P
\end{alltt}

Notice that {\coq} uses the abreviation ``~\citecoq{\exsym\,$x$:$A$, $B$}~''
for \linebreak ``~\citecoq{ex (fun $x$:$A$ \funarrow{} $B$)}~''.


\noindent The former quantifier inhabits the universe of propositions.
As for the conjunction and disjunction connectives, there is also another
version of existential quantification inhabiting the universe $\Set$,
which is noted \texttt{sig $P$}. The syntax
``~\citecoq{\{$x$:$A$ | $B$\}}~'' is an abreviation for ``~\citecoq{sig (fun $x$:$A$ {\funarrow} $B$)}~''.



%\paragraph{The logical connectives.} Conjuction and disjuction are 
%also introduced as recursive types:
%\begin{alltt}
%Print or.
%\end{alltt}
%begin{alltt}
%Print and.
%\end{alltt}


\subsection{Mutually Dependent Definitions}
\label{MutuallyDependent}

Mutually dependent definitions of recursive types are also allowed in
{\coq}.  A typical example of these kind of declaration is the
introduction of the trees of unbounded (but finite) width:
\label{Forest}
\begin{alltt} 
Inductive tree(A:Set)   : Set :=
    node : A {\arrow} forest A \arrow{} tree A 
with  forest (A: Set)   : Set := 
    nochild  : forest A |
    addchild : tree A \arrow{} forest A \arrow{} forest A.
\end{alltt}
\noindent  Yet another example of mutually dependent types are the
predicates \texttt{even} and \texttt{odd} on natural numbers:
\label{Even}
\begin{alltt} 
Inductive 
  even    : nat\arrow{}Prop :=
    evenO : even  O |
    evenS : {\prodsym} n, odd n \arrow{} even (S n)
with
  odd    : nat\arrow{}Prop :=
    oddS : {\prodsym} n, even n \arrow{} odd (S n).
\end{alltt}

\begin{alltt}
Lemma odd_49 : odd (7 * 7).
 simpl; repeat constructor.
Qed.
\end{alltt}



\section{Case Analysis and Pattern-matching}
\label{CaseAnalysis}
\subsection{Non-dependent Case Analysis}
An \textsl{elimination rule} for the type $A$ is some way to use an
object $a:A$ in order to define an object in some  type $B$. 
A natural elimination for an inductive type is \emph{case analysis}.


For instance, any value of type {\nat} is built using either \texttt{O} or \texttt{S}.
Thus, a systematic way of building a value of type $B$ from any 
value of type {\nat} is to associate to \texttt{O} a constant $t_O:B$ and
to every term of the form ``~\texttt{S $p$}~'' a term $t_S:B$. The following
construction has type $B$:
\begin{alltt}
match \(n\) return \(B\) with O \funarrow \(t\sb{O}\) | S p \funarrow \(t\sb{S}\) end
\end{alltt}


In most of the cases, {\coq} is able to infer the type $B$ of the object
defined, so the ``\texttt{return $B$}'' part can be omitted.

The computing rules associated with this construct are the expected ones 
(the notation $t_S\{q/\texttt{p}\}$ stands for the substitution of $p$ by
$q$ in $t_S$:)

\begin{eqnarray*}
\texttt{match $O$ return $b$ with O {\funarrow} $t_O$ | S p {\funarrow} $t_S$ end} &\Longrightarrow& t_O\\
\texttt{match $S\;q$  return $b$ with O {\funarrow} $t_O$ | S p {\funarrow} $t_S$ end}  &\Longrightarrow& t_S\{q/\texttt{p}\}
\end{eqnarray*}


\subsubsection{Example: the predecessor function.}\label{firstpred}
An example of a definition by case analysis is the function which
computes the predecessor of any given natural number:
\begin{alltt}
Definition pred (n:nat) := match n with
                                   | O {\funarrow} O 
                                   | S m {\funarrow} m 
                           end.

Eval simpl in pred 56.
\it{}    = 55
     : nat
\tt
Eval simpl in pred 0.
\it{}    = 0
     : nat

\tt{}Eval simpl in fun p {\funarrow} pred (S p).
\it{}     = fun p : nat {\funarrow} p
     : nat {\arrow} nat
\end{alltt}

As in functional programming, tuples and wild-cards can be used in
patterns \refmancite{Section \ref{ExtensionsOfCases}}. Such
definitions are automatically compiled by {\coq} into an expression which
may contain several nested case expressions. For example, the 
exclusive \emph{or} on booleans can be defined as follows:
\begin{alltt}
Definition xorb (b1 b2:bool) :=
 match b1, b2 with 
 | false, true {\funarrow} true
 | true, false {\funarrow} true
 | _ , _       {\funarrow} false
 end.
\end{alltt}

This kind of definition is compiled in {\coq} as follows\footnote{{\coq} uses
the conditional ``~\citecoq{if $b$ then $a$ else $b$}~'' as an abreviation to
``~\citecoq{match $b$ with true \funarrow{} $a$ | false \funarrow{} $b$ end}~''.}:

\begin{alltt}
Print xorb.
xorb = 
fun b1 b2 : bool {\funarrow}
if b1 then if b2 then false else true 
      else if b2 then true else false
     : bool {\arrow} bool {\arrow} bool
\end{alltt}

\subsection{Dependent Case Analysis}
\label{DependentCase}

For a pattern matching construct of the form
``~\citecoq{match n with \dots end}~'' a more general typing rule
is obtained considering that the type of the whole expression
may also depend on \texttt{n}.
  For instance, let us consider some function 
$Q:\texttt{nat}\arrow{}\texttt{Set}$, and $n:\citecoq{nat}$.
In order to build a term of type $Q\;n$, we can associate
to the constructor \texttt{O} some term $t_O: Q\;\texttt{O}$ and to
the pattern ``~\texttt{S p}~''  some term $t_S : Q\;(S\;p)$.
Notice that the terms $t_O$ and $t_S$ do not have the same type.

The syntax of the \emph{dependent case analysis} and its
associated typing rule make precise how the resulting
type depends on the argument of the pattern matching, and
which constraint holds on the branches of the pattern matching:

\label{Prod-sup-rule}
\[
\begin{array}[t]{l}
Q: \texttt{nat}{\arrow}\texttt{Set}\quad{t_O}:{{Q\;\texttt{O}}}  \quad
\smalljuge{p:\texttt{nat}}{t_p}{{Q\;(\texttt{S}\;p)}} \quad n:\texttt{nat} \\
\hline
{\texttt{match \(n\) as \(n\sb{0}\) return \(Q\;n\sb{0}\) with | O \funarrow \(t\sb{O}\) | S p \funarrow \(t\sb{S}\) end}}:{{Q\;n}}
\end{array}
\]


The interest of this rule of \textsl{dependent} pattern-matching is
that it can also be read as the following logical principle (replacing \citecoq{Set}
by \texttt{Prop} in the type of $Q$): in order to prove
that a property $Q$ holds for all $n$, it is sufficient to prove that
$Q$ holds for {\Z} and that for all $p:\nat$, $Q$ holds for
$(\SUCC\;p)$.  The former, non-dependent version of case analysis can
be obtained from this latter rule just taking $Q$ as a constant
function on $n$.

Notice that destructuring $n$ into \citecoq{O} or ``~\citecoq{S p}~''
 doesn't
make appear in the goal the equalities ``~$n=\citecoq{O}$~''
 and ``~$n=\citecoq{S p}$~''.
They are ``internalized'' in the rules above (see section~\ref{inversion}.)

\subsubsection{Example: strong specification of the predecessor function.}

In Section~\ref{firstpred}, the predecessor function was defined directly
as a function from \texttt{nat} to \texttt{nat}. It remains to prove
that this function has some desired properties. Another way to proceed
is to, first introduce a specification of what is the predecessor of a 
natural number, under the form of a {\coq} type, then build an inhabitant 
of this type: in other words, a realization of this specification. This way, the correctness
of this realization is ensured by {\coq}'s type system.

A reasonable specification for $\pred$ is to say that for all $n$
there exists another $m$ such that either $m=n=0$, or $(\SUCC\;m)$
is equal to  $n$. The function $\pred$ should be just the way to
compute such an $m$. 

\begin{alltt}
Definition pred_spec (n:nat) := 
   \{m:nat | n=0{\coqand} m=0 {\coqor} n = S m\}.
  
Definition  predecessor : {\prodsym} n:nat, pred_spec n.
 intro n; case n.
\it{}  
  n : nat
  ============================
   pred_spec 0

\tt{} unfold pred_spec;exists 0;auto.
\it{}
 =========================================
 {\prodsym} n0 : nat, pred_spec (S n0)
\tt{}
 unfold pred_spec; intro n0; exists n0; auto.
Defined.
\end{alltt}

If we print the term built by {\coq}, we can observe its dependent pattern-matching structure:

\begin{alltt}
predecessor =  fun n : nat {\funarrow}
\textbf{match n as n0 return (pred_spec n0) with}
\textbf{| O {\funarrow}}
    exist (fun m : nat {\funarrow} 0 = 0 {\coqand} m = 0 {\coqor} 0 = S m) 0
      (or_introl (0 = 1) 
                 (conj (refl_equal 0) (refl_equal 0)))
\textbf{| S n0 {\funarrow}}
    exist (fun m : nat {\funarrow} S n0 = 0 {\coqand} m = 0 {\coqor} S n0 = S m) n0
      (or_intror (S n0 = 0 {\coqand} n0 = 0) (refl_equal (S n0)))
\textbf{end}  : {\prodsym} n : nat, \textbf{pred_spec n}
\end{alltt}


Notice that there are many variants to the pattern ``~\texttt{intros \dots; case \dots}~''. Look at the reference manual and/or the book: tactics
\texttt{destruct}, ``~\texttt{intro \emph{pattern}}~'', etc.

\noindent The command \texttt{Extraction} \refmancite{Section
\ref{ExtractionIdent}} can be used to see the computational
contents associated to the \emph{certified} function \texttt{predecessor}:
\begin{alltt}
Extraction predecessor.
\it
(** val predecessor : nat {\arrow} pred_spec **)

let predecessor = function
  | O {\arrow} O
  | S n0 {\arrow} n0
\end{alltt}


\begin{exercise} \label{expand}
Prove the following theorem:
\begin{alltt}
Theorem nat_expand : {\prodsym} n:nat, 
      n = match n with 
                  | 0 {\funarrow} 0 
                  | S p {\funarrow} S p 
          end.
\end{alltt}
\end{exercise}

\subsection{Some Examples of Case Analysis}
\label{CaseScheme}
The reader will find in the Reference manual all details about
typing case analysis (chapter 4: Calculus of Inductive Constructions,
and chapter 15: Extended Pattern-Matching).

The following commented examples will show the different situations to consider.


%\subsubsection{General Scheme}

%Case analysis is then the most basic elimination rule that {\coq}
%provides for inductive  types.  This rule follows a general schema,
%valid for any inductive type $I$. First, if $I$ has type
%``~$\forall\,(z_1:A_1)\ldots(z_r:A_r),S$~'', with $S$ either $\Set$, $\Prop$ or
%$\Type$, then a case expression on $p$ of type ``~$R\;a_1\ldots a_r$~''
% inhabits ``~$Q\;a_1\ldots a_r\;p$~''. The types of the branches of the case expression
%are obtained from the definition of the type in this way: if the type
%of the $i$-th constructor $c_i$  of $R$ is 
%``~$\forall\, (x_1:T_1)\ldots
%(x_n:T_n),(R\;q_1\ldots q_r)$~'', then the  $i-th$ branch must have the
%form ``~$c_i\; x_1\; \ldots \;x_n\;  \funarrow{}\; t_i$~'' where
%$$(x_1:T_1),\ldots, (x_n:T_n) \vdash t_i : Q\;q_1\ldots q_r)$$
% for non-dependent case
%analysis, and $$(x_1:T_1)\ldots (x_n:T_n)\vdash t_i :Q\;q_1\ldots
%q_r\;({c}_i\;x_1\;\ldots x_n)$$ for dependent one. In the
%following section, we illustrate this general scheme for different
%recursive types.
%%\textbf{A vérifier}

\subsubsection{The Empty Type}

In a definition by case analysis, there is one branch for each
introduction rule of the type. Hence, in a definition by case analysis
on $p:\False$ there are no cases to be considered. In other words, the
rule of (non-dependent) case analysis for the type $\False$ is 
(for $s$ in  \texttt{Prop}, \texttt{Set} or \texttt{Type}):

\begin{center}
\snregla {\JM{Q}{s}\;\;\;\;\;
          \JM{p}{\False}}
         {\JM{\texttt{match $p$ return $Q$ with end}}{Q}}
\end{center}

As a corollary, if we could construct an object in $\False$, then it
could be possible to define an object in any type.  The tactic
\texttt{contradiction} \refmancite{Section \ref{Contradiction}}
corresponds to the application of the elimination rule above.  It
searches in the context for an absurd hypothesis (this is, a
hypothesis whose type is $\False$) and then proves the goal by a case
analysis of it.

\begin{alltt}
Theorem fromFalse : False \arrow{} 0=1.
 intro H. 
 contradiction.
Qed.
\end{alltt}


In {\coq} the negation is defined as follows :

\begin{alltt}
Definition not (P:Prop) := P {\arrow} False
\end{alltt}

The proposition ``~\citecoq{not $A$}~'' is also written ``~$\neg A$~''.

If $A$ and $B$ are propositions, $a$ is a proof of $A$ and
$H$ is a proof of $\neg A$,
the term ``~\citecoq{match $H\;a$ return $B$ with end}~'' is a proof term of
$B$.
Thus, if your goal is $B$ and you have some hypothesis $H:\neg A$,
the tactic ``~\citecoq{case $H$}~'' generates a new subgoal with
statement $A$, as shown by  the following example\footnote{Notice that
$a\coqdiff b$ is just an abreviation for ``~\coqnot a= b~''}.

\begin{alltt}
Fact Nosense : 0 {\coqdiff} 0 {\arrow} 2 = 3.
Proof.
  intro H; case H.
\it
===========================
  0 = 0
\tt
  reflexivity.
Qed.
\end{alltt}

The tactic ``~\texttt{absurd $A$}~'' (where $A$ is any proposition), 
is based on the same principle, but
generates two subgoals: $A$ and $\neg A$, for solving $B$.

\subsubsection{The Equality Type}

Let $A:\Type$, $a$, $b$ of type $A$, and $\pi$ a proof of 
$a=b$. Non dependent case analysis of $\pi$ allows us to
associate to any proof  of ``~$Q\;a$~'' a proof of ``~$Q\;b$~'',
where $Q:A\arrow{} s$ (where $s\in\{\Prop, \Set, \Type\}$).
The following term is a proof  of ``~$Q\;a \arrow{} Q\;b$~''.

\begin{alltt}
fun H : Q a {\funarrow}
  match \(\pi\) in (_ = y) return Q y with
     refl_equal {\funarrow} H
  end
\end{alltt}
Notice the header of the \texttt{match} construct.
It expresses how the resulting type ``~\citecoq{Q y}~'' depends on 
the \emph{type} of \texttt{p}.
Notice also that in the pattern introduced by the keyword \texttt{in},
the parameter \texttt{a} in the type ``~\texttt{a = y}~'' must be
implicit, and replaced by a wildcard '\texttt{\_}'.


Therefore, case analysis on a proof of the equality $a=b$
amounts to replacing all the occurrences of the term $b$ with the term
$a$ in the goal to be proven. Let us illustrate this through an
example: the transitivity property of this equality. 
\begin{alltt}
Theorem trans : {\prodsym} n m p:nat, n=m \arrow{} m=p \arrow{} n=p.
Proof.
 intros n m p eqnm.  
\it{}  
  n : nat
  m : nat
  p : nat
  eqnm : n = m
  ============================
   m = p {\arrow} n = p
\tt{} case eqnm.
\it{}
  n : nat
  m : nat
  p : nat
  eqnm : n = m
  ============================
   n = p {\arrow} n = p
\tt{} trivial.
Qed.
\end{alltt}

%\noindent The case analysis on the hypothesis $H:n=m$ yields the
%tautological subgoal $n=p\rightarrow n=p$, that is directly proven by
%the tactic \texttt{Trivial}.

\begin{exercise}
Prove the symmetry property of  equality.
\end{exercise}

Instead of using \texttt{case}, we can use the tactic 
\texttt{rewrite} \refmancite{Section \ref{Rewrite}}.  If $H$ is a proof
of  $a=b$, then
``~\citecoq{rewrite $H$}~''
 performs a case analysis on a proof of $b=a$, obtained by applying a
symmetry theorem to $H$. This application of symmetry allows us to rewrite
the equality from left to right, which looks more natural. An optional
parameter (either \texttt{\arrow{}} or \texttt{$\leftarrow$}) can be used to precise
in which sense the equality must be rewritten.  By default,
``~\texttt{rewrite} $H$~'' corresponds to ``~\texttt{rewrite \arrow{}} $H$~''
\begin{alltt}
Lemma Rw :  {\prodsym} x y: nat, y = y * x {\arrow} y * x * x = y.
 intros x y e; do 2 rewrite <- e.
\it
1 subgoal
  
  x : nat
  y : nat
  e : y = y * x
  ============================
   y = y
\tt
 reflexivity.
Qed.
\end{alltt}

Notice that, if $H:a=b$, then the tactic ``~\texttt{rewrite $H$}~''
 replaces \textsl{all} the
occurrences of $a$ by $b$. However, in certain situations we could be
interested in rewriting some of the occurrences, but not all of them.
This can be done using the tactic \texttt{pattern} \refmancite{Section
\ref{Pattern}}.  Let us consider yet another example to
illustrate this.

Let us start with some simple theorems of arithmetic; two of them 
are already proven in the Standard Library, the last is left as an exercise.

\begin{alltt}
\it
mult_1_l
     : {\prodsym} n : nat, 1 * n = n

mult_plus_distr_r
     : {\prodsym} n m p : nat, (n + m) * p = n * p + m * p

mult_distr_S : {\prodsym} n p : nat, n * p + p = (S n)* p.
\end{alltt}

Let us now prove a simple result:

\begin{alltt}
Lemma four_n : {\prodsym} n:nat, n+n+n+n = 4*n.
Proof.
 intro n;rewrite <- (mult_1_l n).
\it
  n : nat
  ============================
   1 * n + 1 * n + 1 * n + 1 * n = 4 * (1 * n)
\end{alltt}

We can see that the \texttt{rewrite} tactic call replaced \emph{all}
the occurrences of \texttt{n} by the term ``~\citecoq{1 * n}~''.
If we want to do the rewriting ony on the leftmost occurrence of
\texttt{n}, we can mark this occurrence using the \texttt{pattern}
tactic:


\begin{alltt}
 Undo.
 intro n; pattern n at 1.
 \it
 n : nat
  ============================
 (fun n0 : nat {\funarrow} n0 + n + n + n = 4 * n) n
\end{alltt}
Applying the tactic ``~\citecoq{pattern  n at 1}~'' allowed us
to explicitly abstract the first occurrence of \texttt{n} from the
goal, putting this goal under the form ``~\citecoq{$Q$ n}~'',
thus pointing to \texttt{rewrite} the particular predicate on $n$
that we search to prove. 


\begin{alltt}
 rewrite <- mult_1_l.
\it
1 subgoal
  
  n : nat
  ============================
   1 * n + n + n + n = 4 * n
\tt
 repeat rewrite   mult_distr_S.
\it
  n : nat
  ============================
   4 * n = 4 * n
\tt
 trivial.
Qed.
\end{alltt}

\subsubsection{The Predicate $n {\leq} m$}


The last but one instance of the elimination schema that we will illustrate is
case analysis for the predicate $n {\leq} m$:

Let $n$ and $p$ be terms of type \citecoq{nat}, and $Q$ a predicate 
of type $\citecoq{nat}\arrow{}\Prop$.
If $H$ is a proof of ``~\texttt{n {\coqle} p}~'',
$H_0$ a proof of ``~\texttt{$Q$  n}~'' and
$H_S$ a proof of ``~\citecoq{{\prodsym}m:nat, n {\coqle}  m {\arrow} Q (S m)}~'',
then the term
\begin{alltt}
match H in (_ {\coqle} q) return (Q q) with
    | le_n {\funarrow} H0
    | le_S m Hm {\funarrow} HS m Hm
end
\end{alltt}
 is a proof term of ``~\citecoq{$Q$ $p$}~''.


The two patterns of this \texttt{match} construct describe
all possible forms of proofs of ``~\citecoq{n {\coqle} m}~'' (notice
again that the general parameter \texttt{n} is implicit in
 the ``~\texttt{in \dots}~''
clause and is absent from the match patterns.


Notice that the choice of introducing some of the arguments of the
predicate as being general parameters in its definition has
consequences on the rule of case analysis that is derived. In
particular, the type $Q$ of the object defined by the case expression
only depends on the indexes of the predicate, and not on the general
parameters.  In the definition of the predicate $\leq$, the first
argument of this relation is a general parameter of the
definition. Hence, the predicate $Q$ to be proven only depends on the
second argument of the relation. In other words, the integer $n$ is
also a general parameter of the rule of case analysis.

An example of an application of this rule is the following theorem,
showing that any integer greater or equal than $1$ is the successor of another
natural number:

\begin{alltt}
Lemma predecessor_of_positive : 
 {\prodsym} n, 1 {\coqle} n {\arrow} {\exsym} p:nat, n = S p.
Proof.
 intros n H;case H.
\it
  n : nat
  H : 1 {\coqle} n
  ============================
   {\exsym} p : nat, 1 = S p
\tt
  exists 0; trivial.
\it

  n : nat
  H : 1 {\coqle} n
  ============================
   {\prodsym} m : nat, 0 {\coqle} m {\arrow} {\exsym} p : nat, S m = S p
\tt
  intros m _  .
  exists m.
  trivial.
Qed.
\end{alltt}


\subsubsection{Vectors}

The \texttt{vector} polymorphic and dependent family of types will
give an idea of the most general scheme of pattern-matching.

For instance, let us define a function for computing the tail of
any vector. Notice that we shall build a \emph{total} function,
by considering that the tail of an empty vector is this vector itself.
In that sense, it will be slightly different from the \texttt{Vtail}
function of the Standard Library, which is defined only for vectors
of type ``~\citecoq{vector $A$ (S $n$)}~''.

The header of the function we want to build is the following:

\begin{verbatim}
Definition Vtail_total 
   (A : Set) (n : nat) (v : vector A n) : vector A (pred n):=
\end{verbatim}

Since the branches will not have the same type
(depending on the parameter \texttt{n}),
the body of this function is a dependent pattern matching on 
\citecoq{v}.
So we will have :
\begin{verbatim}
match v in (vector _ n0) return (vector A (pred n0)) with
\end{verbatim}

The first branch  deals with the constructor \texttt{Vnil} and must
return a value in ``~\citecoq{vector A (pred 0)}~'', convertible
to ``~\citecoq{vector A 0}~''. So, we propose:
\begin{alltt}
| Vnil {\funarrow} Vnil A
\end{alltt}

The second branch considers a vector in ``~\citecoq{vector A (S n0)}~''
of the form
``~\citecoq{Vcons A n0 v0}~'', with ``~\citecoq{v0:vector A n0}~'',
and must return a value in ``~\citecoq{vector A (pred (S n0))}~'',
convertible to ``~\citecoq{v0:vector A n0}~''.
This second branch is thus :
\begin{alltt}
| Vcons _ n0 v0 {\funarrow} v0
\end{alltt}

Here is the full definition:

\begin{alltt}
Definition Vtail_total 
   (A : Set) (n : nat) (v : vector A n) : vector A (pred n):=
match v in (vector _ n0) return (vector A (pred n0)) with
| Vnil {\funarrow} Vnil A
| Vcons _ n0 v0 {\funarrow} v0
end.
\end{alltt}


\subsection{Case Analysis and Logical Paradoxes}

In the previous section we have illustrated the general scheme for
generating the rule of case analysis associated to some recursive type
from the definition of the type. However, if the logical soundness is
to be preserved, certain restrictions to this schema are
necessary. This section provides a brief explanation of these
restrictions.


\subsubsection{The Positivity Condition}
\label{postypes}

In order to make sense of recursive types as types closed under their
introduction rules, a constraint has to be imposed on the possible
forms of such rules. This constraint, known as the
\textsl{positivity condition}, is necessary to prevent the user from
naively introducing some recursive types which would open the door to
logical paradoxes.  An example of such a dangerous type is the
``inductive type'' \citecoq{Lambda}, whose only constructor is 
\citecoq{lambda} of type \citecoq{(Lambda\arrow False)\arrow Lambda}.
 Following the pattern
given in Section \ref{CaseScheme}, the rule of (non dependent) case
analysis for \citecoq{Lambda} would be the following:

\begin{center}
\snregla {\JM{Q}{\Prop}\;\;\;\;\;
          \JM{p}{\texttt{Lambda}}\;\;\;\;\;
          {h : {\texttt{Lambda}}\arrow\False\; \vdash\; t\,:\,Q}}
         {\JM{\citecoq{match $p$ return $Q$ with lambda h {\funarrow} $t$  end}}{Q}}
\end{center}

In order to avoid paradoxes, it is impossible to construct
the type \citecoq{Lambda} in {\coq}:

\begin{alltt}
Inductive Lambda : Set :=
  lambda : (Lambda {\arrow} False) {\arrow} Lambda. 
\it
Error: Non strictly positive occurrence of "Lambda" in
 "(Lambda {\arrow} False) {\arrow} Lambda"
\end{alltt}

In order to explain this danger, we
will declare some constants for simulating the construction of 
\texttt{Lambda} as an inductive type.

Let us open some section, and declare two variables, the first one for
\texttt{Lambda}, the other for the constructor \texttt{lambda}.

\begin{alltt}
Section Paradox.
Variable Lambda : Set.
Variable lambda : (Lambda {\arrow} False) {\arrow}Lambda.
\end{alltt}

Since \texttt{Lambda} is not a truely inductive type, we can't use
the \texttt{match} construct. Nevertheless, we can simulate it by a
variable \texttt{matchL} such that
``~\citecoq{matchL $l$  $Q$ (fun $h$ : Lambda {\arrow} False {\funarrow} $t$)}~''
should be understood  as 
``~\citecoq{match $l$ return $Q$ with | lambda h {\funarrow} $t$)}~''


\begin{alltt}
Variable matchL : Lambda {\arrow} 
                  {\prodsym} Q:Prop, ((Lambda {\arrow}False) {\arrow} Q) {\arrow}
                  Q.
\end{alltt}

From these constants, it is possible to define application by case
analysis. Then, through auto-application, the well-known looping term
$(\lambda x.(x\;x)\;\lambda x.(x\;x))$ provides a proof of falsehood.

\begin{alltt}
Definition application (f x: Lambda) :False :=
  matchL f False (fun h {\funarrow} h x).

Definition Delta :  Lambda := 
  lambda (fun x : Lambda {\funarrow} application x x).

Definition loop : False := application Delta Delta.

Theorem two_is_three : 2 = 3.
Proof.
 elim loop.
Qed.

End Paradox.
\end{alltt}

\noindent This example can be seen as a formulation of Russell's
paradox in type theory associating $(\textsl{application}\;x\;x)$ to the
formula $x\not\in x$, and \textsl{Delta} to the set $\{ x \mid
x\not\in x\}$. If \texttt{matchL} would satisfy the reduction rule
associated to case analysis, that is, 
$$ \citecoq{matchL (lambda $f$) $Q$ $h$} \Longrightarrow h\;f$$
then the term \texttt{loop}
would compute into itself.  This is not actually surprising, since the
proof of the logical soundness of {\coq} strongly lays on the property
that any well-typed term must terminate. Hence, non-termination is
usually a synonymous of inconsistency.

%\paragraph{} In this case, the construction of a non-terminating
%program comes from the so-called \textsl{negative occurrence} of
%$\Lambda$ in the type of the constructor $\lambda$. In order to be
%admissible for {\coq}, all the occurrences of the recursive type in its
%own introduction rules must be positive, in the sense on the following
%definition:
%
%\begin{enumerate}
%\item $R$ is positive in $(R\;\vec{t})$;
%\item $R$ is positive in $(x: A)C$ if it does not 
%occur in $A$ and $R$ is positive in $C$; 
%\item if $P\equiv (\vec{x}:\vec{T})Q$, then $R$ is positive in $(P
%\rightarrow C)$ if $R$ does not occur in $\vec{T}$, $R$ is positive
%in $C$, and either
%\begin{enumerate} 
%\item $Q\equiv (R\;\vec{q})$ or 
%\item $Q\equiv (J\;\vec{t})$, \label{relax}
%      where $J$ is a recursive type, and for any term $t_i$ either : 
%   \begin{enumerate}
%      \item $R$ does not occur in $t_i$, or
%      \item $t_i\equiv (z:\vec{Z})(R\;\vec{q})$, $R$ does not occur
%       in $\vec{Z}$, $t_i$ instantiates a general
%       parameter of $J$, and this parameter is positive in the 
%       arguments of the constructors of $J$. 
%   \end{enumerate}
%\end{enumerate}
%\end{enumerate}
%\noindent Those types obtained by erasing option (\ref{relax}) in the
%definition above are called \textsl{strictly positive} types.


\paragraph{} In this case, the construction of a non-terminating
program comes from the so-called \textsl{negative occurrence} of
\texttt{Lambda} in the argument of the constructor \texttt{lambda}. 

The reader will find in the Reference Manual a complete formal 
definition of the notions of \emph{positivity condition} and
\emph{strict positivity} that an inductive definition must satisfy.


%In order to be
%admissible for {\coq}, the type $R$ must be positive in the types of the
%arguments of its own introduction rules, in the sense on the following
%definition:

%\textbf{La définition du manuel de référence est plus complexe:
%la recopier ou donner seulement des exemples?
%}
%\begin{enumerate}
%\item $R$ is positive in $T$ if $R$ does not occur in $T$;
%\item $R$ is positive in $(R\;\vec{t})$ if $R$ does not occur in $\vec{t}$;
%\item $R$ is positive in $(x:A)C$ if it does not
%      occur in $A$ and $R$ is positive in $C$; 
%\item $R$ is positive in $(J\;\vec{t})$, \label{relax}
%      if $J$ is a recursive type, and for any term $t_i$ either : 
%   \begin{enumerate}
%      \item $R$ does not occur in $t_i$, or
%      \item $R$ is positive in $t_i$,  $t_i$ instantiates a general
%       parameter of $J$, and this parameter is positive in the 
%       arguments of the constructors of $J$. 
%   \end{enumerate}
%\end{enumerate}

%\noindent When we can show that $R$ is positive without using the item
%(\ref{relax}) of the definition above, then we say that $R$ is
%\textsl{strictly positive}.

%\textbf{Changer le discours sur les ordinaux}

Notice that the positivity condition does not forbid us to
put  functional recursive
arguments in the constructors. 

For instance, let us consider the type of infinitely branching trees,
with labels in \texttt{Z}.
\begin{alltt}
Require Import ZArith.

Inductive itree : Set :=
| ileaf : itree
| inode : Z {\arrow} (nat {\arrow} itree) {\arrow} itree.
\end{alltt}

In this representation, the $i$-th child of a tree 
represented by ``~\texttt{inode $z$ $s$}~'' is obtained by applying
the function $s$ to $i$.
The following definitions show how to construct a tree with a single 
node, a tree of height 1 and a tree of height 2:

\begin{alltt}
Definition isingle l := inode l (fun i {\funarrow} ileaf).

Definition t1 := inode 0 (fun n {\funarrow} isingle (Z_of_nat n)).

Definition t2 := 
 inode 0 
      (fun n : nat {\funarrow} 
                   inode (Z_of_nat n)
                   (fun p {\funarrow} isingle (Z_of_nat (n*p)))).
\end{alltt}


Let us define a preorder on infinitely branching trees.
 In order to compare two non-leaf trees,
it is necessary to compare each of their children 
 without taking care of the order in which they
appear:

\begin{alltt}
Inductive itree_le : itree{\arrow} itree {\arrow} Prop :=
  | le_leaf : {\prodsym} t, itree_le  ileaf t
  | le_node : {\prodsym} l l' s s', 
                Zle l l' {\arrow} 
                ({\prodsym} i, {\exsym} j:nat, itree_le (s i) (s' j)){\arrow} 
                itree_le  (inode  l s) (inode  l' s').

\end{alltt}

Notice that a call to the predicate \texttt{itree\_le} appears as
a general parameter of the inductive type  \texttt{ex} (see Sect.\ref{ex-def}).
This kind of definition is accepted by {\coq}, but may lead to some
difficulties, since the induction principle automatically 
generated by the system
is not the most appropriate (see chapter 14 of~\cite{coqart} for a detailed
explanation).


The following definition, obtained by 
skolemising the
proposition \linebreak $\forall\, i,\exists\, j,(\texttt{itree\_le}\;(s\;i)\;(s'\;j))$ in
the type of \texttt{itree\_le}, does not present this problem:


\begin{alltt} 
Inductive itree_le' : itree{\arrow} itree {\arrow} Prop :=
  | le_leaf'  : {\prodsym} t, itree_le'  ileaf t
  | le_node' : {\prodsym} l l' s s' g, 
                  Zle l l' {\arrow}  
                  ({\prodsym} i, itree_le' (s i) (s' (g i))) {\arrow} 
                  itree_le'  (inode  l s) (inode  l' s').

\end{alltt}
\iffalse
\begin{alltt}
Lemma t1_le'_t2 :  itree_le' t1 t2.
Proof.
 unfold t1, t2.
 constructor 2  with (fun i : nat {\funarrow} 2 * i).
 auto with zarith.
 unfold isingle;
 intro i ; constructor 2 with (fun i :nat {\funarrow} i).
 auto with zarith.
 constructor .
Qed.
\end{alltt}
\fi

%In general, strictly positive definitions are preferable to only
%positive ones. The reason is that it is sometimes difficult to derive
%structural induction combinators for the latter ones. Such combinators
%are automatically generated for strictly positive types, but not for
%the only positive ones. Nevertheless, sometimes non-strictly positive
%definitions provide a smarter or shorter way of declaring a recursive
%type.

Another example is the type of trees 
 of unbounded width, in which a recursive subterm 
\texttt{(ltree A)} instantiates the type  of polymorphic lists:

\begin{alltt} 
Require Import List.

Inductive ltree  (A:Set) : Set :=  
          lnode   : A {\arrow} list (ltree A) {\arrow} ltree A.
\end{alltt}

This declaration can be transformed  
adding an extra type to the definition, as was done in Section
\ref{MutuallyDependent}.


\subsubsection{Impredicative Inductive Types}

An inductive type $R$ inhabiting a universe $S$ is \textsl{predicative}
if the introduction rules of $R$ do not make a universal
quantification on a universe containing $S$. All the recursive types
previously introduced are examples of predicative types. An example of
an impredicative one is the following type:
%\textsl{exT}, the dependent product
%of a certain set (or proposition) $x$, and a proof of a property $P$
%about $x$.

%\begin{alltt} 
%Print exT.
%\end{alltt}
%\textbf{ttention, EXT c'est ex!}
%\begin{alltt}
%Check (exists P:Prop, P {\arrow} not P).  
%\end{alltt}

%This type is useful for expressing existential quantification over
%types, like ``there exists a proposition $x$ such that $(P\;x)$''
%---written $(\textsl{EXT}\; x:Prop \mid (P\;x))$ in {\coq}. However,

\begin{alltt}
Inductive prop : Prop :=
 prop_intro : Prop {\arrow} prop.
\end{alltt}

Notice
that the constructor of this type can be used to inject any
proposition --even itself!-- into the type. A careless use of such a
self-contained objects may lead to a variant of Burali-Forti's
paradox. The construction of Burali-Forti's paradox is more
complicated than Russel's one, so we will not describe it here, and
point the interested reader to \cite{Bar98,Coq86}.


\begin{alltt}
Lemma prop_inject: prop.
Proof prop_intro prop.
\end{alltt}

Another example is the second order existential quantifier for propositions:

\begin{alltt}
Inductive ex_Prop  (P : Prop {\arrow} Prop) : Prop :=
  exP_intro : {\prodsym} X : Prop, P X {\arrow} ex_Prop P.
\end{alltt}

%\begin{alltt}
%(*
%Check (match prop_inject with (prop_intro p _) {\funarrow} p end).

%Error: Incorrect elimination of "prop_inject" in the inductive type
% ex
%The elimination predicate ""fun _ : prop {\funarrow} Prop" has type
% "prop {\arrow} Type"
%It should be one of :
% "Prop"

%Elimination of an inductive object of sort : "Prop"
%is not allowed on a predicate in sort : "Type"
%because non-informative objects may not construct informative ones.

%*)
%Print prop_inject.

%(*
%prop_inject = 
%prop_inject = prop_intro prop (fun H : prop {\funarrow} H)
%     : prop
%*)
%\end{alltt}

% \textbf{Et par ça?
%}

Notice that predicativity on sort \citecoq{Set} forbids us to build
the following definitions.


\begin{alltt}
Inductive aSet : Set :=
  aSet_intro: Set {\arrow} aSet.

\it{}User error: Large non-propositional inductive types must be in Type
\tt 
Inductive ex_Set  (P : Set {\arrow} Prop) : Set :=
  exS_intro : {\prodsym} X : Set, P X {\arrow} ex_Set P.

\it{}User error: Large non-propositional inductive types must be in Type
\end{alltt}

Nevertheless, one can define types like \citecoq{aSet} and \citecoq{ex\_Set}, as inhabitants of \citecoq{Type}.

\begin{alltt}
Inductive ex_Set  (P : Set {\arrow} Prop) : Type :=
  exS_intro : {\prodsym} X : Set, P X {\arrow} ex_Set P.
\end{alltt}

In the following example, the inductive type \texttt{typ} can be defined,
but the term associated with the interactive Definition of
\citecoq{typ\_inject} is incompatible with {\coq}'s hierarchy of universes:


\begin{alltt}
Inductive  typ : Type := 
  typ_intro : Type {\arrow} typ. 

Definition typ_inject: typ.
 split; exact typ.
\it Proof completed
\tt{}Defined.
\it
Error: Universe Inconsistency.
\tt
Abort.
\end{alltt}

One possible way of avoiding this new source of paradoxes is to
restrict the kind of eliminations by case analysis that can be done on
impredicative types. In particular, projections on those universes
equal or bigger than the one inhabited by the impredicative type must
be forbidden \cite{Coq86}. A consequence of this restriction is that it
is not possible to define the first projection of the type
``~\citecoq{ex\_Prop $P$}~'':
\begin{alltt}
Check (fun (P:Prop{\arrow}Prop)(p: ex_Prop P) {\funarrow}
      match p with exP_intro X HX {\funarrow} X end).
\it
Error:
Incorrect elimination of "p" in the inductive type  
"ex_Prop", the return type has sort "Type" while it should be 
"Prop"

Elimination of an inductive object of sort "Prop"
is not allowed on a predicate in sort "Type"
because proofs can be eliminated only to build proofs.
\end{alltt}

%In order to explain why, let us consider for example the following
%impredicative type \texttt{ALambda}.
%\begin{alltt} 
%Inductive ALambda : Set := 
%          alambda : (A:Set)(A\arrow{}False)\arrow{}ALambda.
%
%Definition Lambda : Set := ALambda.
%Definition lambda : (ALambda\arrow{}False)\arrow{}ALambda := (alambda ALambda).
%Lemma CaseAL : (Q:Prop)ALambda\arrow{}((ALambda\arrow{}False)\arrow{}Q)\arrow{}Q.
%\end{alltt}
%
%This type contains all the elements of the dangerous type $\Lambda$
%described at the beginning of this section.  Try to construct the
%non-ending term $(\Delta\;\Delta)$ as an object of
%\texttt{ALambda}. Why is it not possible?

\subsubsection{Extraction Constraints}

There is a final constraint on case analysis that is not motivated by
the potential introduction of paradoxes, but for compatibility reasons
with {\coq}'s extraction mechanism \refmancite{Appendix
\ref{CamlHaskellExtraction}}. This mechanism is based on the
classification of basic types into the universe $\Set$ of sets and the
universe $\Prop$ of propositions.  The objects of a type in the
universe $\Set$ are considered as relevant for computation
purposes. The objects of a type in $\Prop$ are considered just as
formalised comments, not necessary for execution. The extraction
mechanism consists in erasing such formal comments in order to obtain
an executable program. Hence, in general, it is not possible to define
an object in a set (that should be kept by the extraction mechanism)
by case analysis of a proof (which will be thrown away).

Nevertheless, this general rule has an exception which is important in
practice: if the definition proceeds by case analysis on a proof of a
\textsl{singleton proposition} or an empty type (\emph{e.g.} \texttt{False}),
 then it is allowed. A singleton
proposition is a non-recursive proposition with a single constructor
$c$, all whose arguments are proofs. For example, the propositional
equality and the conjunction of two propositions are examples of
singleton propositions.

%From the point of view of the extraction
%mechanism, such types are isomorphic to a type containing a single
%object $c$, so a definition $\Case{x}{c \Rightarrow b}$ is 
%directly replaced by $b$ as an extra optimisation.

\subsubsection{Strong Case Analysis on Proofs}

One could consider allowing 
 to define a proposition $Q$ by case
analysis on the proofs of another recursive proposition $R$. As we
will see in Section \ref{Discrimination}, this would enable one to prove that
different introduction rules of $R$ construct different
objects. However, this property would be in contradiction with the principle
of excluded middle of classical logic, because this principle entails
that the proofs of a proposition cannot be distinguished. This
principle is not provable in {\coq}, but it is frequently introduced by
the users as an axiom, for reasoning in classical logic. For this
reason, the definition of propositions by case analysis on proofs is
 not allowed in {\coq}.

\begin{alltt}

Definition comes_from_the_left (P Q:Prop)(H:P{\coqor}Q): Prop :=
 match H with
         |  or_introl p {\funarrow} True 
         |  or_intror q {\funarrow} False
 end.
\it
Error:
Incorrect elimination of "H" in the inductive type  
"or", the return type has sort "Type" while it should be 
"Prop"

Elimination of an inductive object of sort "Prop"
is not allowed on a predicate in sort "Type"
because proofs can be eliminated only to build proofs.

\end{alltt}

On the other hand, if we replace the proposition $P {\coqor} Q$ with
the informative type $\{P\}+\{Q\}$, the elimination  is accepted:

\begin{alltt}
Definition comes_from_the_left_sumbool
            (P Q:Prop)(x:\{P\} + \{Q\}): Prop :=
  match x with
         |  left  p {\funarrow} True 
         |  right q {\funarrow} False
  end.
\end{alltt}


\subsubsection{Summary of Constraints}

To end with this section, the following table summarizes which
universe $U_1$ may inhabit an object of type $Q$ defined by case
analysis on $x:R$, depending on the universe $U_2$ inhabited by the
inductive types $R$.\footnote{In the box indexed by $U_1=\citecoq{Type}$
and $U_2=\citecoq{Set}$, the answer ``yes''  takes into account the 
predicativity of sort \citecoq{Set}. If you are working with the
option ``impredicative-set'', you must put in this box the
condition ``if $R$ is predicative''.}


\begin{center}
\renewcommand{\multirowsetup}{\centering} \newlength{\LL}
\settowidth{\LL}{$x : R : s_1$}
\begin{tabular}{|c|c|c|c|c|}
\hline
\multirow{5}{\LL}{$x : R : U_2$} &
\multicolumn{4}{|c|}{$Q : U_1$}\\
\hline
&                 &\textsl{Set}      & \textsl{Prop} & \textsl{Type}\\
\cline{2-5}
&\textsl{Set}     &  yes             &   yes         &  yes\\
\cline{2-5}
&\textsl{Prop}    & if $R$ singleton &   yes         &  no\\
\cline{2-5}
&\textsl{Type}    &  yes             &   yes         &  yes\\
\hline
\end{tabular}
\end{center}

\section{Some Proof Techniques Based on Case Analysis}
\label{CaseTechniques}

In this section we illustrate the use of case analysis as a proof
principle, explaining the proof techniques behind three very useful
{\coq} tactics, called \texttt{discriminate}, \texttt{injection} and
\texttt{inversion}. 

\subsection{Discrimination of introduction rules}
\label{Discrimination}

In the informal semantics of recursive types described in Section
\ref{Introduction} it was said that each of the introduction rules of a
recursive type is considered as being different from all the others. 
It is possible to capture this fact inside the logical system using
the propositional equality. We take as example the following theorem,
stating that \textsl{O} constructs a natural number different 
from any of those constructed with \texttt{S}. 

\begin{alltt}
Theorem S_is_not_O : {\prodsym} n, S n {\coqdiff} 0. 
\end{alltt}

In order to prove this theorem, we first define a proposition by case
analysis on natural numbers, so that the proposition is true for {\Z}
and false for any natural number constructed with {\SUCC}. This uses
the empty and singleton type introduced in Sections \ref{Introduction}.

\begin{alltt}
Definition Is_zero (x:nat):= match x with 
                                     | 0 {\funarrow} True  
                                     | _ {\funarrow} False
                             end.
\end{alltt}

\noindent Then, we prove the following lemma:

\begin{alltt}
Lemma O_is_zero : {\prodsym} m, m = 0 {\arrow} Is_zero m.
Proof.
  intros m H; subst m. 
\it{}
================
 Is_zero 0
\tt{}
simpl;trivial.
Qed.
\end{alltt}

\noindent Finally, the proof of \texttt{S\_is\_not\_O} follows by the
application of the previous lemma to $S\;n$.


\begin{alltt}

 red; intros n Hn.
 \it{}   
  n : nat
  Hn : S n = 0
  ============================
   False \tt

 apply O_is_zero with (m := S n).
 assumption.
Qed.
\end{alltt}


The tactic \texttt{discriminate} \refmancite{Section \ref{Discriminate}} is
a special-purpose tactic for proving disequalities between two
elements of a recursive type introduced by different constructors. It
generalizes the proof method described here for natural numbers to any
[co]-inductive type. This tactic is also capable of proving disequalities
where the difference is not in the constructors at the head of the
terms, but deeper inside them. For example, it can be used to prove
the following theorem:

\begin{alltt}
Theorem disc2 : {\prodsym} n, S (S n) {\coqdiff} 1. 
Proof.
 intros n Hn; discriminate.
Qed.
\end{alltt}

When there is an assumption $H$ in the context stating a false
equality $t_1=t_2$, \texttt{discriminate} solves the goal by first
proving $(t_1\not =t_2)$ and then reasoning by absurdity with respect
to $H$:

\begin{alltt}
Theorem disc3 : {\prodsym} n, S (S n) = 0 {\arrow} {\prodsym} Q:Prop, Q.
Proof.
 intros n Hn Q.
 discriminate.
Qed.
\end{alltt}

\noindent In this case, the proof proceeds by absurdity with respect
to the false equality assumed, whose negation is proved by
discrimination.

\subsection{Injectiveness of introduction rules}

Another useful property about recursive types is the
\textsl{injectiveness} of introduction rules, i.e., that whenever two
objects were built using the same introduction rule, then this rule
should have been applied to the same element. This can be stated
formally using the propositional equality:

\begin{alltt}
Theorem inj : {\prodsym} n m, S n = S m {\arrow} n = m.
Proof.
\end{alltt}

\noindent This theorem is just a corollary of a lemma about the
predecessor function:

\begin{alltt}
 Lemma inj_pred : {\prodsym} n m, n = m {\arrow} pred n = pred m.
 Proof.
  intros n m eq_n_m.
  rewrite eq_n_m.
  trivial.
 Qed.
\end{alltt}
\noindent Once this lemma is proven, the theorem follows directly
from it:
\begin{alltt}
 intros n m eq_Sn_Sm.
 apply inj_pred with (n:= S n) (m := S m); assumption.
Qed.
\end{alltt}

This proof method is implemented by the tactic \texttt{injection}
\refmancite{Section \ref{injection}}. This tactic is applied to
a term $t$ of type ``~$c\;{t_1}\;\dots\;t_n = c\;t'_1\;\dots\;t'_n$~'', where $c$ is some constructor of
an inductive type. The tactic \texttt{injection} is applied as deep as 
possible to derive the equality of all pairs of subterms of $t_i$ and $t'_i$
placed in the same position. All these equalities are put as antecedents 
of the current goal.



Like \texttt{discriminate}, the tactic \citecoq{injection} 
can be also applied if $x$ does not
occur in a direct sub-term, but somewhere deeper inside it. Its
application may leave some trivial goals that can be easily solved
using the tactic \texttt{trivial}.

\begin{alltt}

 Lemma list_inject : {\prodsym} (A:Set)(a b :A)(l l':list A),
             a :: b :: l = b :: a :: l' {\arrow} a = b {\coqand} l = l'.
Proof.
 intros A a b l l' e.


\it
  e : a :: b :: l = b :: a :: l'
  ============================
   a = b {\coqand} l = l'
\tt
 injection e.
\it
  ============================
   l = l' {\arrow} b = a {\arrow} a = b {\arrow} a = b {\coqand} l = l'

\tt{} auto.
Qed.
\end{alltt}

\subsection{Inversion Techniques}\label{inversion}

In section \ref{DependentCase}, we motivated the rule of dependent case
analysis as a way of internalizing the informal equalities $n=O$ and
$n=(\SUCC\;p)$ associated to each case. This internalisation
consisted in instantiating $n$ with the corresponding term in the type
of each branch. However, sometimes it could be better to internalise
these equalities as extra hypotheses --for example, in order to use
the tactics \texttt{rewrite}, \texttt{discriminate} or
\texttt{injection} presented in the previous sections. This is
frequently the case when the element analysed is denoted by a term
which is not a variable, or when it is an object of a particular
instance of a recursive family of types. Consider for example the
following theorem:

\begin{alltt}
Theorem not_le_Sn_0 : {\prodsym} n:nat, ~ (S n {\coqle} 0).
\end{alltt}

\noindent Intuitively, this theorem should follow by case analysis on
the hypothesis $H:(S\;n\;\leq\;\Z)$, because no introduction rule allows
to instantiate the arguments of \citecoq{le} with respectively a successor
and zero. However, there
is no way of capturing this with the typing rule for case analysis
presented in section \ref{Introduction}, because it does not take into
account what particular instance of the family the type of $H$ is.
Let us try it:
\begin{alltt}
Proof.
 red; intros n H; case H.
\it 2 subgoals
  
  n : nat
  H : S n {\coqle} 0
  ============================
   False

subgoal 2 is:
 {\prodsym} m : nat, S n {\coqle} m {\arrow} False
\tt
Undo.
\end{alltt}

\noindent What is necessary here is to make available the equalities
``~$\SUCC\;n = \Z$~'' and ``~$\SUCC\;m = \Z$~''
 as extra hypotheses of the
branches, so that the goal can be solved using the
\texttt{Discriminate} tactic. In order to obtain the desired
equalities as hypotheses, let us prove an auxiliary lemma, that our
theorem is a corollary of:

\begin{alltt}
 Lemma not_le_Sn_0_with_constraints :
  {\prodsym} n p , S n {\coqle} p {\arrow}  p = 0 {\arrow} False.
  Proof.
   intros n p H; case H .
\it
2 subgoals
  
  n : nat
  p : nat
  H : S n {\coqle} p
  ============================
   S n = 0 {\arrow} False

subgoal 2 is:
 {\prodsym} m : nat, S n {\coqle} m {\arrow} S m = 0 {\arrow} False
\tt
 intros;discriminate.
 intros;discriminate.
Qed.
\end{alltt}
\noindent Our main theorem can now be solved by an application of this lemma:
\begin{alltt}
Show.
\it
2 subgoals
  
  n : nat
  p : nat
  H : S n {\coqle} p
  ============================
   S n = 0 {\arrow} False

subgoal 2 is:
 {\prodsym} m : nat, S n {\coqle} m {\arrow} S m = 0 {\arrow} False
\tt
 eapply not_le_Sn_0_with_constraints; eauto.
Qed.
\end{alltt}


The general method to address such situations consists in changing the
goal to be proven into an implication, introducing as preconditions
the equalities needed to eliminate the cases that make no
sense. This proof technique is implemented by the tactic
\texttt{inversion} \refmancite{Section \ref{Inversion}}. In order
to prove a goal $G\;\vec{q}$ from an object of type $R\;\vec{t}$,
this tactic automatically generates a lemma $\forall, \vec{x}.
(R\;\vec{x}) \rightarrow \vec{x}=\vec{t}\rightarrow \vec{B}\rightarrow
(G\;\vec{q})$, where the list of propositions $\vec{B}$ correspond to
the subgoals that cannot be directly proven using
\texttt{discriminate}. This lemma can either be saved for later
use, or generated interactively. In this latter case, the subgoals
yielded by the tactic are the hypotheses $\vec{B}$ of the lemma. If the
lemma has been stored, then the tactic \linebreak
 ``~\citecoq{inversion \dots using \dots}~'' can be
used to apply it. 

Let us show both techniques on our previous example:

\subsubsection{Interactive mode}

\begin{alltt}
Theorem not_le_Sn_0' : {\prodsym} n:nat, ~ (S n {\coqle} 0).
Proof.
 red; intros n H ; inversion H.
Qed.
\end{alltt}


\subsubsection{Static mode}

\begin{alltt}

Derive Inversion le_Sn_0_inv with ({\prodsym} n :nat, S n {\coqle}  0).
Theorem le_Sn_0'' : {\prodsym} n p : nat, ~ S n {\coqle} 0 .
Proof.
 intros n p H; 
 inversion H using le_Sn_0_inv.
Qed.
\end{alltt}


In the example above, all the cases are solved using discriminate, so
there remains no subgoal to be proven (i.e. the list $\vec{B}$ is
empty). Let us present a second example, where this list is not empty:


\begin{alltt}
TTheorem le_reverse_rules : 
     {\prodsym} n m:nat, n {\coqle} m {\arrow} 
                     n = m {\coqor}  
                     {\exsym} p, n {\coqle}  p {\coqand} m = S p.
Proof.
 intros n m H; inversion H.
\it
2 subgoals



  
  n : nat
  m : nat
  H : n {\coqle} m
  H0 : n = m
  ============================
   m = m {\coqor} ({\exsym} p : nat, m {\coqle} p {\coqand} m = S p)

subgoal 2 is:
 n = S m0 {\coqor} ({\exsym} p : nat, n {\coqle} p {\coqand} S m0 = S p)
\tt
 left;trivial.
 right; exists m0; split; trivial.
\it
Proof completed
\end{alltt}

This example shows how this tactic can be used to ``reverse'' the
introduction rules of a recursive type, deriving the possible premises
that could lead to prove a given instance of the predicate. This is
why these tactics are called \texttt{inversion} tactics: they go back
from conclusions to premises.

The hypotheses corresponding to the propositional equalities are not
needed in this example, since the tactic does the necessary rewriting
to solve the subgoals.  When the equalities are no longer needed after
the inversion, it is better to use the tactic
\texttt{Inversion\_clear}. This variant of the tactic clears from the
context all the equalities introduced.

\begin{alltt}
Restart.
 intros n m H; inversion_clear H.
\it
\it
  
  n : nat
  m : nat
  ============================
   m = m {\coqor} ({\exsym} p : nat, m {\coqle} p {\coqand} m = S p)
\tt
 left;trivial.
\it
  n : nat
  m : nat
  m0 : nat
  H0 : n {\coqle} m0
  ============================
   n = S m0 {\coqor} ({\exsym} p : nat, n {\coqle} p {\coqand} S m0 = S p)
\tt
 right; exists m0; split; trivial.
Qed.
\end{alltt}


%This proof technique works in most of the cases, but not always.  In
%particular, it could not if the list $\vec{t}$ contains a term $t_j$
%whose type $T$ depends on a previous term $t_i$, with $i<j$. Remark
%that if this is the case, the propositional equality $x_j=t_j$ is not
%well-typed, since $x_j:T(x_i)$ but $t_j:T(t_i)$, and both types are
%not convertible (otherwise, the problem could be solved using the
%tactic \texttt{Case}).



\begin{exercise}
Consider the following language of arithmetic expression, and 
its operational semantics, described by a set of rewriting rules.
%\textbf{J'ai enlevé une règle de commutativité de l'addition qui 
%me paraissait bizarre du point de vue de la sémantique opérationnelle}

\begin{alltt}
Inductive ArithExp : Set :=
   | Zero : ArithExp 
   | Succ : ArithExp {\arrow} ArithExp
   | Plus : ArithExp {\arrow} ArithExp {\arrow} ArithExp.

Inductive RewriteRel : ArithExp {\arrow} ArithExp {\arrow} Prop :=
  |  RewSucc  : {\prodsym} e1 e2 :ArithExp,
                  RewriteRel e1 e2 {\arrow}
                   RewriteRel (Succ e1) (Succ e2) 
  |  RewPlus0 : {\prodsym} e:ArithExp,
                  RewriteRel (Plus Zero e) e 
  |  RewPlusS : {\prodsym} e1 e2:ArithExp,
                  RewriteRel e1 e2 {\arrow}
                  RewriteRel (Plus (Succ e1) e2) 
                             (Succ (Plus e1 e2)).

\end{alltt}
\begin{enumerate}
\item Prove that \texttt{Zero} cannot be rewritten any further.
\item Prove that an expression of the form ``~$\texttt{Succ}\;e$~'' is always 
rewritten
into an expression of the same form.
\end{enumerate}
\end{exercise}

%Theorem zeroNotCompute : (e:ArithExp)~(RewriteRel Zero e).
%Intro e.
%Red.
%Intro H.
%Inversion_clear H.
%Defined.
%Theorem evalPlus : 
%  (e1,e2:ArithExp)
%    (RewriteRel (Succ e1) e2)\arrow{}(EX e3 : ArithExp | e2=(Succ e3)).
%Intros e1 e2 H.
%Inversion_clear H.
%Exists e3;Reflexivity.
%Qed.


\section{Inductive Types and Structural Induction} 
\label{StructuralInduction}

Elements of inductive types  are well-founded with
respect to the structural order induced by the constructors of the
type. In addition to case analysis, this extra hypothesis about
well-foundedness justifies a stronger elimination rule for them, called
\textsl{structural induction}.  This form of elimination consists in
defining a value ``~$f\;x$~'' from some element $x$ of the inductive type
$I$, assuming that values have been already associated in the same way
to the sub-parts of $x$ of type $I$.


Definitions by structural induction are expressed through the
\texttt{Fixpoint} command \refmancite{Section
\ref{Fixpoint}}. This command is quite close to the
\texttt{let-rec} construction of functional programming languages.
For example, the following definition introduces the addition of two
natural numbers (already defined in the Standard Library:)

\begin{alltt} 
Fixpoint plus (n p:nat) \{struct n\} : nat :=
  match n with
          | 0 {\funarrow} p
          | S m {\funarrow} S (plus m p)
 end.
\end{alltt}

The definition is by structural induction on the first argument of the
function. This is indicated  by the ``~\citecoq{\{struct n\}}~''
directive in the function's header\footnote{This directive is optional
in the case of a function of a single argument}.
  In
order to be accepted, the definition must satisfy a syntactical
condition, called the \textsl{guardedness condition}. Roughly
speaking, this condition constrains the arguments of a recursive call
to be pattern variables, issued from a case analysis of the formal
argument of the function pointed by the \texttt{struct} directive.
 In the case of the
function \texttt{plus}, the argument \texttt{m} in the recursive call is a
pattern variable issued from a case analysis of \texttt{n}. Therefore, the
definition is accepted.

Notice that we could have defined the addition with structural induction 
on its second argument:
\begin{alltt} 
Fixpoint plus' (n p:nat) \{struct p\} : nat :=
    match p with
          | 0 {\funarrow} n
          | S q {\funarrow} S (plus' n q)
    end.
\end{alltt}

%This notation is useful when defining a function whose decreasing
%argument has a dependent type. As an example, consider the following
%recursivly defined proof of the theorem
%$(n,m:\texttt{nat})n<m \rightarrow (S\;n)<(S\;m)$:
%\begin{alltt} 
%Fixpoint lt_n_S [n,m:nat;p:(lt n m)] : (lt (S n) (S m)) :=
% <[n0:nat](lt (S n) (S n0))>
%     Cases p of
%       lt_intro1        {\funarrow} (lt_intro1 (S n))
%    | (lt_intro2 m1 p2) {\funarrow} (lt_intro2 (S n) (S m1) (lt_n_S n m1 p2))
%     end.
%\end{alltt}

%The guardedness condition must be satisfied only by the last argument
%of the enclosed list. For example, the following declaration is an
%alternative way of defining addition:

%\begin{alltt}
%Reset add.
%Fixpoint add [n:nat] : nat\arrow{}nat :=
%  Cases n of 
%       O    {\funarrow} [x:nat]x
%    | (S m) {\funarrow} [x:nat](add m (S x))
%  end.
%\end{alltt}

In the following definition of addition, 
the second argument of \verb@plus''@ grows at each
recursive call. However, as the first one always decreases, the
definition is sound.
\begin{alltt}
Fixpoint plus'' (n p:nat) \{struct n\} : nat :=
 match n with
          | 0 {\funarrow} p
          | S m {\funarrow} plus'' m (S p)
 end.
\end{alltt}

 Moreover, the argument in the recursive call
could be a deeper component of $n$.  This is the case in the following
definition of a boolean function determining whether a number is even
or odd:

\begin{alltt} 
Fixpoint even_test (n:nat) : bool :=
  match n 
  with 0 {\funarrow}  true
     | 1 {\funarrow}  false
     | S (S p) {\funarrow} even_test p
  end.
\end{alltt}

Mutually dependent definitions by structural induction are also
allowed. For example, the previous function \textsl{even} could alternatively
be defined using an auxiliary function \textsl{odd}:

\begin{alltt}
Reset even_test.



Fixpoint even_test (n:nat) : bool :=
  match n 
  with 
      | 0 {\funarrow}  true
      | S p {\funarrow} odd_test p
  end
with odd_test (n:nat) : bool :=
  match n
  with 
     | 0 {\funarrow} false
     | S p {\funarrow} even_test p
 end.
\end{alltt}

%\begin{exercise}
%Define a function by structural induction that computes the number of
%nodes of a tree structure defined in page \pageref{Forest}.
%\end{exercise}

Definitions by structural induction are computed 
 only when they are applied, and the decreasing argument
is a term having a constructor at the head. We can check this using
the \texttt{Eval} command, which computes the normal form of a well
typed term.

\begin{alltt}
Eval simpl in even_test.
\it
    = even_test
     : nat {\arrow} bool
\tt 
Eval simpl in (fun x : nat {\funarrow} even x).
\it
     = fun x : nat {\funarrow} even x
     : nat {\arrow} Prop
\tt
Eval simpl in (fun x : nat => plus 5 x).
\it
     =  fun x : nat {\funarrow} S (S (S (S (S x))))

\tt
Eval simpl in (fun x : nat {\funarrow} even_test (plus 5 x)).
\it
    = fun x : nat {\funarrow} odd_test x
     : nat {\arrow} bool
\tt
Eval simpl in (fun x : nat {\funarrow} even_test (plus x 5)).
\it
    = fun x : nat {\funarrow} even_test (x + 5)
     : nat {\arrow} bool
\end{alltt}
 

%\begin{exercise}
%Prove that the second definition of even satisfies the following
%theorem:
%\begin{verbatim}
%Theorem unfold_even : 
% (x:nat)
%   (even x)= (Cases x of 
%                 O        {\funarrow} true
%              | (S O)     {\funarrow} false 
%              | (S (S m)) {\funarrow} (even m)
%              end).
%\end{verbatim}
%\end{exercise}

\subsection{Proofs by Structural Induction}

The principle of structural induction can be also used in order to
define proofs, that is, to prove theorems. Let us call an
\textsl{elimination combinator} any function that, given a predicate
$P$, defines a proof of ``~$P\;x$~'' by structural induction on $x$.  In
{\coq}, the principle of proof by induction on natural numbers is a
particular case of an elimination combinator. The definition of this
combinator depends on three general parameters: the predicate to be
proven, the base case, and the inductive step:

\begin{alltt}
Section Principle_of_Induction.
Variable    P               : nat {\arrow} Prop.
Hypothesis  base_case       : P 0.
Hypothesis  inductive_step  : {\prodsym} n:nat, P n {\arrow} P (S n).
Fixpoint nat_ind  (n:nat)   : (P n) := 
   match n return P n with
          | 0 {\funarrow} base_case
          | S m {\funarrow} inductive_step m (nat_ind m)
   end. 

End Principle_of_Induction.
\end{alltt}

As this proof principle is used very often, {\coq} automatically generates it
when an inductive type is introduced.  Similar principles
\texttt{nat\_rec} and \texttt{nat\_rect} for defining objects in the
universes $\Set$ and $\Type$ are also automatically generated
\footnote{In fact, whenever possible, {\coq} generates the
principle \texttt{$I$\_rect}, then derives from it the
weaker principles  \texttt{$I$\_ind} and \texttt{$I$\_rec}.
If some principle has to be defined by hand, the user may try
to build \texttt{$I$\_rect} (if possible). Thanks to {\coq}'s conversion
rule, this principle can be used directly to build proofs and/or
programs.}. The
command \texttt{Scheme} \refmancite{Section \ref{Scheme}} can be
used to generate an elimination combinator from certain parameters,
like the universe that the defined objects must inhabit, whether the
case analysis in the definitions must be dependent or not, etc. For
example, it can be used to generate an elimination combinator for
reasoning on even natural numbers from the mutually dependent
predicates introduced in page \pageref{Even}. We do not display the
combinators here by lack of space, but you can see them using the
\texttt{Print} command.

\begin{alltt}
Scheme Even_induction := Minimality for even Sort Prop
with   Odd_induction  := Minimality for odd  Sort Prop.
\end{alltt}

\begin{alltt}
Theorem even_plus_four : {\prodsym} n:nat, even n {\arrow} even (4+n).
Proof.
 intros n H.
 elim H using Even_induction with (P0 := fun n {\funarrow} odd (4+n));
 simpl;repeat constructor;assumption.
Qed.
\end{alltt}

Another example of an elimination combinator is the principle 
of double induction on natural numbers, introduced by the following
definition:

\begin{alltt}
Section Principle_of_Double_Induction.
Variable    P              : nat {\arrow} nat {\arrow}Prop.
Hypothesis  base_case1     : {\prodsym} m:nat, P 0 m.
Hypothesis  base_case2     : {\prodsym} n:nat, P (S n) 0.
Hypothesis  inductive_step : {\prodsym} n m:nat, P n m {\arrow}
                                       \,\, P (S n) (S m).

Fixpoint nat_double_ind (n m:nat)\{struct n\} : P n m := 
 match n, m return P n m with 
 |     0 ,    x   {\funarrow}  base_case1 x 
 |  (S x),    0   {\funarrow} base_case2 x
 |  (S x), (S y) {\funarrow} inductive_step x y (nat_double_ind x y)
 end.
End Principle_of_Double_Induction.
\end{alltt}

Changing the type of $P$ into $\nat\rightarrow\nat\rightarrow\Set$,
another combinator \texttt{nat\_double\_rec} for constructing 
(certified) programs can be defined in exactly the same way.
This definition is left as an exercise.\label{natdoublerec}

\iffalse
\begin{alltt}
Section Principle_of_Double_Recursion.
Variable    P               : nat {\arrow} nat {\arrow} Set.
Hypothesis  base_case1      : {\prodsym} x:nat, P 0 x.
Hypothesis  base_case2      : {\prodsym} x:nat, P (S x) 0.
Hypothesis  inductive_step   : {\prodsym} n m:nat, P n m {\arrow} P (S n) (S m).
Fixpoint nat_double_rec (n m:nat)\{struct n\} : P n m := 
  match n, m return P n m with 
            0 ,     x   {\funarrow}  base_case1 x 
         |  (S x),    0   {\funarrow} base_case2 x
         |  (S x), (S y) {\funarrow} inductive_step x y (nat_double_rec x y)
     end.
End Principle_of_Double_Recursion.
\end{alltt}
\fi
For instance the function computing the minimum of two natural
numbers can be defined in the following way:

\begin{alltt}
Definition min : nat {\arrow} nat {\arrow} nat  := 
  nat_double_rec (fun (x y:nat) {\funarrow} nat)
                 (fun (x:nat) {\funarrow} 0)
                 (fun (y:nat) {\funarrow} 0)
                 (fun (x y r:nat) {\funarrow} S r).
Eval compute in (min 5 8).
\it
= 5 : nat
\end{alltt}


%\begin{exercise}
%
%Define the combinator \texttt{nat\_double\_rec}, and apply it
%to give another definition of \citecoq{le\_lt\_dec} (using the theorems
%of the \texttt{Arith} library).
%\end{exercise}

\subsection{Using Elimination Combinators.} 
The tactic \texttt{apply} can be used to apply one of these proof
principles during the development of a proof. 

\begin{alltt}
Lemma not_circular : {\prodsym} n:nat, n {\coqdiff} S n.
Proof.
 intro n.
 apply nat_ind with (P:= fun n {\funarrow} n {\coqdiff} S n).
\it



2 subgoals
  
  n : nat
  ============================
   0 {\coqdiff} 1


subgoal 2 is:
 {\prodsym} n0 : nat, n0 {\coqdiff} S n0 {\arrow} S n0 {\coqdiff} S (S n0)

\tt
 discriminate.
 red; intros n0 Hn0 eqn0Sn0;injection eqn0Sn0;trivial.
Qed.
\end{alltt}

The tactic \texttt{elim} \refmancite{Section \ref{Elim}} is a
refinement of \texttt{apply}, specially designed for the application
of elimination combinators.  If $t$ is an object of an inductive type
$I$, then ``~\citecoq{elim $t$}~'' tries to find an abstraction $P$ of the
current goal $G$ such that $(P\;t)\equiv G$. Then it solves the goal
applying ``~$I\texttt{\_ind}\;P$~'', where $I$\texttt{\_ind} is the
combinator associated to $I$.  The different cases of the induction
then appear as subgoals that remain to be solved.
In the previous proof, the tactic call ``~\citecoq{apply nat\_ind with (P:= fun n {\funarrow} n {\coqdiff} S n)}~'' can simply be replaced with ``~\citecoq{elim n}~''.

The option ``~\citecoq{\texttt{elim} $t$ \texttt{using} $C$}~''
 allows to use a
derived combinator $C$ instead of the default one. Consider the
following theorem, stating that equality is decidable on natural
numbers:

\label{iseqpage}
\begin{alltt}
Lemma eq_nat_dec : {\prodsym} n p:nat, \{n=p\}+\{n {\coqdiff} p\}.
Proof.
 intros n p.
\end{alltt}

Let us prove this theorem using the combinator \texttt{nat\_double\_rec}
of section~\ref{natdoublerec}. The example also illustrates how
\texttt{elim} may sometimes fail in finding a suitable abstraction $P$
of the goal. Note that if ``~\texttt{elim n}~''
 is used directly on the
goal, the result is not the expected one.

\vspace{12pt}

%\pagebreak
\begin{alltt}
 elim n using nat_double_rec.
\it
4 subgoals
  
  n : nat
  p : nat
  ============================
   {\prodsym} x : nat, \{x = p\} + \{x {\coqdiff} p\}

subgoal 2 is:
 nat {\arrow} \{0 = p\} + \{0 {\coqdiff} p\}

subgoal 3 is:
 nat {\arrow} {\prodsym} m : nat, \{m = p\} + \{m {\coqdiff} p\} {\arrow} \{S m = p\} + \{S m {\coqdiff} p\}

subgoal 4 is:
 nat
\end{alltt}

The four sub-goals obtained do not correspond to the premises that
would be expected for the principle \texttt{nat\_double\_rec}. The
problem comes from the fact that 
this principle for eliminating $n$
has a universally quantified formula as conclusion, which confuses
\texttt{elim} about the right way of abstracting the goal. 

%In effect, let us consider the type of the goal before the call to
%\citecoq{elim}: ``~\citecoq{\{n = p\} + \{n {\coqdiff} p\}}~''.

%Among all the abstractions that can be built by ``~\citecoq{elim n}~''
%let us consider this one
%$P=$\citecoq{fun n :nat {\funarrow} fun q : nat {\funarrow} {\{q= p\} + \{q {\coqdiff} p\}}}.
%It is easy to verify that 
%$P$ has type \citecoq{nat {\arrow} nat {\arrow} Set}, and that, if some 
%$q:\citecoq{nat}$ is given, then $P\;q\;$ matches the current goal.
%Then applying \citecoq{nat\_double\_rec} with $P$ generates
%four goals, corresponding to 
 



Therefore,
in this case the abstraction must be explicited using the tactic
\texttt{pattern}. Once the right abstraction is provided, the rest of
the proof is immediate:

\begin{alltt}
Undo.
 pattern p,n.
\it
  n : nat
  p : nat
  ============================
   (fun n0 n1 : nat {\funarrow} \{n1 = n0\} + \{n1 {\coqdiff} n0\}) p n
\tt
 elim n using nat_double_rec.
\it
3 subgoals
  
  n : nat
  p : nat
  ============================
   {\prodsym} x : nat, \{x = 0\} + \{x {\coqdiff} 0\}

subgoal 2 is:
 {\prodsym} x : nat, \{0 = S x\} + \{0 {\coqdiff} S x\}
subgoal 3 is:
 {\prodsym} n0 m : nat, \{m = n0\} + \{m {\coqdiff} n0\} {\arrow} \{S m = S n0\} + \{S m {\coqdiff} S n0\}

\tt
 destruct x; auto.
 destruct x; auto.
 intros n0 m H; case H.
 intro eq; rewrite eq ; auto.
 intro neg; right; red ; injection 1; auto.
Defined.
\end{alltt}


Notice that the tactic ``~\texttt{decide equality}~''
\refmancite{Section\ref{DecideEquality}} generalises the proof
above to a large class of inductive types.  It can be used for proving
a proposition of the form 
$\forall\,(x,y:R),\{x=y\}+\{x{\coqdiff}y\}$, where $R$ is an inductive datatype
all whose constructors take informative arguments ---like for example
the type {\nat}:

\begin{alltt}
Definition eq_nat_dec' : {\prodsym} n p:nat, \{n=p\} + \{n{\coqdiff}p\}.
 decide equality.
Defined.
\end{alltt}

\begin{exercise}
\begin{enumerate}
\item Define a recursive  function \emph{nat2itree}
mapping any natural number $n$ into an infinitely branching
tree of height $n$.
\item Provide an elimination combinator for these trees.
\item Prove that the relation \citecoq{itree\_le} is a preorder 
(i.e. reflexive and transitive).
\end{enumerate}
\end{exercise}

\begin{exercise} \label{zeroton}
Define the type of lists, and a predicate ``being an ordered list''
using an inductive family. Then, define the function
$(from\;n)=0::1\;\ldots\; n::\texttt{nil}$ and prove that it always generates an
ordered list.
\end{exercise}


\subsection{Well-founded Recursion}
\label{WellFoundedRecursion}

Structural induction is a strong elimination rule for inductive types.
This method can be used to define any function whose termination is
based on the well-foundedness of certain order relation $R$ decreasing
at each recursive call. What makes this principle so strong is the
possibility of reasoning by structural induction on the proof that
certain $R$ is well-founded.  In order to illustrate this we have
first to introduce the predicate of accessibility.

\begin{alltt}
Print Acc.
\it
Inductive Acc (A : Set) (R : A {\arrow} A {\arrow} Prop) (x:A) : Prop :=
    Acc_intro : ({\prodsym} y : A, R y x {\arrow} Acc R y) {\arrow} Acc R x
For Acc: Argument A is implicit
For Acc_intro: Arguments A, R are implicit

\dots
\end{alltt}

\noindent This inductive predicate characterize those elements $x$ of
$A$ such that any descending $R$-chain $\ldots x_2\;R\;x_1\;R\;x$
starting from $x$ is finite. A well-founded relation is a relation
such that all the elements of $A$ are accessible.  

Consider now the problem of representing in {\coq} the following ML
function $\textsl{div}(x,y)$ on natural numbers, which computes
$\lceil\frac{x}{y}\rceil$ if $y>0$ and yields $x$ otherwise.

\begin{verbatim}
let rec div x y = 
  if x = 0 then 0
  else if y = 0 then x
       else (div (x-y) y)+1;;
\end{verbatim}


The equality test on natural numbers can be represented as the
function \textsl{eq\_nat\_dec} defined page \pageref{iseqpage}. Giving $x$ and
$y$, this function yields either the value $(\textsl{left}\;p)$ if
there exists a proof $p:x=y$, or the value $(\textsl{right}\;q)$ if
there exists $q:a\not = b$. The subtraction function is already
defined in the library \citecoq{Minus}. 

Hence, direct translation of the ML function \textsl{div} would be:

\begin{alltt}
Require Import Minus.

Fixpoint div (x y:nat)\{struct x\}: nat :=
 if eq_nat_dec x 0 
  then 0
  else if eq_nat_dec y 0
       then x
       else S (div (x-y) y).

\it Error:
Recursive definition of div is ill-formed.
In environment
div : nat {\arrow} nat {\arrow} nat
x : nat
y : nat
_ : x {\coqdiff} 0
_ : y {\coqdiff} 0

Recursive call to div has principal argument equal to
"x - y"
instead of a subterm of x
\end{alltt}


The program \texttt{div} is rejected by {\coq} because it does not verify
the syntactical condition to ensure termination. In particular, the
argument of the recursive call is not a pattern variable issued from a
case analysis on $x$. 
We would have the same problem if we had the directive
``~\citecoq{\{struct y\}}~'' instead of ``~\citecoq{\{struct x\}}~''.
However, we know that this program always
stops. One way to justify its termination is to define it by
structural induction on a proof that $x$ is accessible trough the
relation $<$. Notice that any natural number $x$ is accessible
for this relation. In order to do this, it is first necessary to prove
some auxiliary lemmas, justifying that the first argument of
\texttt{div} decreases at each recursive call.

\begin{alltt}
Lemma minus_smaller_S : {\prodsym} x y:nat, x - y < S x.
Proof.
 intros x y; pattern y, x;
 elim x using nat_double_ind.
 destruct x0; auto with arith.
 simpl; auto with arith.
 simpl; auto with arith.
Qed.


Lemma minus_smaller_positive : 
 {\prodsym} x y:nat, x {\coqdiff}0 {\arrow} y {\coqdiff} 0 {\arrow}  x - y < x.
Proof.
 destruct x; destruct y; 
 ( simpl;intros; apply minus_smaller || 
   intros; absurd (0=0); auto).
Qed.
\end{alltt}

\noindent The last two lemmas are necessary to prove that for any pair
of positive natural numbers $x$ and $y$, if $x$ is accessible with
respect to \citecoq{lt}, then so is $x-y$.

\begin{alltt}
Definition minus_decrease : {\prodsym} x y:nat, Acc lt x {\arrow} 
                                         x {\coqdiff} 0 {\arrow} 
                                         y {\coqdiff} 0 {\arrow}
                                         Acc lt (x-y).
Proof.
 intros x y H; case H.
 intros Hz posz posy. 
 apply Hz; apply minus_smaller_positive; assumption.
Defined.
\end{alltt}

Let us take a look at the proof of the lemma \textsl{minus\_decrease}, since
the way in which it has been proven is crucial for what follows.
\begin{alltt}
Print minus_decrease.
\it
minus_decrease = 
fun (x y : nat) (H : Acc lt x) {\funarrow}
match H in (Acc _ y0) return (y0 {\coqdiff} 0 {\arrow} y {\coqdiff} 0 {\arrow} Acc lt (y0 - y)) with
| Acc_intro z Hz {\funarrow}
    fun (posz : z {\coqdiff} 0) (posy : y {\coqdiff} 0) {\funarrow}
    Hz (z - y) (minus_smaller_positive z y posz posy)
end
     : {\prodsym} x y : nat, Acc lt x {\arrow} x {\coqdiff} 0 {\arrow} y {\coqdiff} 0 {\arrow} Acc lt (x - y)

\end{alltt}
\noindent Notice that the function call 
$(\texttt{minus\_decrease}\;n\;m\;H)$
indeed yields an accessibility proof that is \textsl{structurally
smaller} than its argument $H$, because it is (an application of) its
recursive component $Hz$.  This enables to justify the following
definition of \textsl{div\_aux}:

\begin{alltt}
Definition div_aux (x y:nat)(H: Acc lt x):nat.
 fix 3.
 intros.
 refine (if eq_nat_dec x 0 
         then 0 
         else if eq_nat_dec y 0 
              then y
              else div_aux (x-y) y _).
\it
 div_aux : {\prodsym} x : nat, nat {\arrow} Acc lt x {\arrow} nat
  x : nat
  y : nat
  H : Acc lt x
  _ : x {\coqdiff} 0
  _0 : y {\coqdiff} 0
  ============================
   Acc lt (x - y)

\tt
 apply (minus_decrease x y H);auto. 
Defined.
\end{alltt}

The main division function is easily defined, using the theorem
\citecoq{lt\_wf} of the library \citecoq{Wf\_nat}. This theorem asserts that
\citecoq{nat} is well founded w.r.t. \citecoq{lt}, thus any natural number
is accessible.
\begin{alltt}
Definition div x y := div_aux x y (lt_wf x). 
\end{alltt}

Let us explain the proof above. In the definition of \citecoq{div\_aux},
what decreases is not $x$ but the \textsl{proof} of the accessibility
of $x$. The tactic ``~\texttt{fix 3}~'' is used to indicate that the proof
proceeds by structural induction on the third argument of the theorem
--that is, on the accessibility proof. It also introduces a new
hypothesis in the context, named as the current theorem, and with the
same type as the goal. Then, the proof is refined with an incomplete
proof term, containing a hole \texttt{\_}.  This hole corresponds to the proof
of accessibility for $x-y$, and is filled up with the (smaller!)
accessibility proof provided by the function \texttt{minus\_decrease}. 


\noindent Let us take a look to the term \textsl{div\_aux} defined:

\pagebreak
\begin{alltt}
Print div_aux.
\it
div_aux = 
(fix div_aux (x y : nat) (H : Acc lt x) \{struct H\} : nat :=
   match eq_nat_dec x 0 with
   | left _ {\funarrow} 0
   | right _ {\funarrow}
       match eq_nat_dec y 0 with
       | left _ {\funarrow} y
       | right _0 {\funarrow} div_aux (x - y) y (minus_decrease x y H _ _0)
       end
   end)
     : {\prodsym} x : nat, nat {\arrow} Acc lt x {\arrow} nat

\end{alltt}

If the non-informative parts from this proof --that is, the
accessibility proof-- are erased, then we obtain exactly the program
that we were looking for. 
\begin{alltt}

Extraction div.

\it
let div x y =
  div_aux x y
\tt

Extraction div_aux.

\it
let rec div_aux x y =
  match eq_nat_dec x O with
    | Left {\arrow} O
    | Right {\arrow}
        (match eq_nat_dec y O with
           | Left {\arrow} y
           | Right {\arrow} div_aux (minus x y) y)
\end{alltt}

This methodology enables the representation
of any program whose termination can be proved in {\coq}. Once the
expected properties from this program have been verified, the
justification of its termination can be thrown away, keeping just the
desired computational behavior for it.

\section{A case study in dependent elimination}\label{CaseStudy}

Dependent types are very expressive, but ignoring some useful
techniques can cause some problems to the beginner.
Let us consider again the type of vectors (see section~\ref{vectors}).
We want to prove a quite trivial property: the only value of type
``~\citecoq{vector A 0}~'' is ``~\citecoq{Vnil $A$}~''.

Our first naive attempt leads to a \emph{cul-de-sac}.
\begin{alltt}
Lemma vector0_is_vnil : 
  {\prodsym} (A:Set)(v:vector A 0), v = Vnil A.
Proof.
 intros A v;inversion v.
\it
1 subgoal
  
  A : Set
  v : vector A 0
  ============================
   v = Vnil A
\tt
Abort.
\end{alltt}

Another attempt is to do a case analysis on a vector of any length
$n$, under an explicit hypothesis $n=0$. The tactic 
\texttt{discriminate} will help us to get rid of the case 
$n=\texttt{S $p$}$. 
Unfortunately, even the statement of our lemma is refused!

\begin{alltt}
 Lemma vector0_is_vnil_aux : 
 {\prodsym} (A:Set)(n:nat)(v:vector A n), n = 0 {\arrow} v = Vnil A.

\it
Error: In environment
A : Set
n : nat
v : vector A n
e : n = 0
The term "Vnil A" has type "vector A 0" while it is expected to have type
 "vector A n"
\end{alltt}

In effect, the equality ``~\citecoq{v = Vnil A}~'' is ill typed,
because the type ``~\citecoq{vector A n}~'' is not \emph{convertible}
with ``~\citecoq{vector A 0}~''.

This problem can be solved if we consider the heterogeneous 
equality \citecoq{JMeq} \cite{conor:motive}
which allows us to consider terms of different types, even if this
equality can only be proven for terms in the same type.
The axiom \citecoq{JMeq\_eq}, from the library \citecoq{JMeq} allows us to convert a
heterogeneous equality to a standard one.

\begin{alltt}
Lemma vector0_is_vnil_aux : 
   {\prodsym} (A:Set)(n:nat)(v:vector A n), 
      n= 0 {\arrow} JMeq v (Vnil A).
Proof.
 destruct v.
 auto.
 intro; discriminate.
Qed.
\end{alltt}

Our property of vectors of null length can be easily proven:

\begin{alltt}
Lemma vector0_is_vnil : {\prodsym} (A:Set)(v:vector A 0), v = Vnil A.
 intros a v;apply JMeq_eq.
 apply vector0_is_vnil_aux.
 trivial.
Qed.
\end{alltt}

It is interesting to look at another proof of 
\citecoq{vector0\_is\_vnil}, which illustrates a technique developed
and used by various people (consult in the \emph{Coq-club} mailing
list archive the contributions by Yves Bertot, Pierre Letouzey, Laurent Théry,
Jean Duprat, and Nicolas Magaud, Venanzio Capretta and Conor McBride).
This technique is also used for unfolding  infinite list definitions
(see chapter13 of~\cite{coqart}).
Notice that this definition does not rely on any axiom (\emph{e.g.} \texttt{JMeq\_eq}).

We first give a new definition of the identity on vectors. Before that,
we make  the use of constructors and selectors lighter thanks to
the implicit arguments feature:

\begin{alltt}
Implicit Arguments Vcons [A n].
Implicit Arguments Vnil [A].
Implicit Arguments Vhead [A n].
Implicit Arguments Vtail [A n].

Definition Vid : {\prodsym} (A : Set)(n:nat), vector A n {\arrow} vector A n.
Proof.
 destruct n; intro v.
 exact Vnil.
 exact (Vcons (Vhead v) (Vtail v)).
Defined.
\end{alltt}


Then we prove that \citecoq{Vid} is the identity on vectors:

\begin{alltt}
Lemma Vid_eq : {\prodsym} (n:nat) (A:Set)(v:vector A n), v=(Vid _ n v).
Proof.
 destruct v.

\it
   A : Set
  ============================
   Vnil = Vid A 0 Vnil

subgoal 2 is:
  Vcons a v = Vid A (S n) (Vcons a v)
\tt
 reflexivity.
 reflexivity.
Defined.
\end{alltt}

Why defining a new identity function on vectors? The following
dialogue shows that \citecoq{Vid} has some interesting computational
properties:

\begin{alltt}
Eval simpl in (fun (A:Set)(v:vector A 0) {\funarrow} (Vid _ _ v)).
\it = fun (A : Set) (_ : vector A 0) {\funarrow} Vnil
     : {\prodsym} A : Set, vector A 0 {\arrow} vector A 0

\end{alltt}

Notice that the plain identity on vectors doesn't convert \citecoq{v}
into \citecoq{Vnil}.
\begin{alltt}
Eval simpl in (fun (A:Set)(v:vector A 0) {\funarrow} v).
\it = fun (A : Set) (v : vector A 0) {\funarrow} v
     : {\prodsym} A : Set, vector A 0 {\arrow} vector A 0
\end{alltt}

Then we prove easily that any vector of length 0 is \citecoq{Vnil}:

\begin{alltt}
Theorem zero_nil : {\prodsym} A (v:vector A 0), v = Vnil.
Proof.
 intros.
 change (Vnil (A:=A)) with (Vid _ 0 v). 
\it
1 subgoal
  
  A : Set
  v : vector A 0
  ============================
   v = Vid A 0 v
\tt
 apply Vid_eq.
Defined.
\end{alltt}

A similar result can be proven about vectors of strictly positive
lenght\footnote{As for \citecoq{Vid} and \citecoq{Vid\_eq}, this definition 
is from Jean Duprat.}.

\begin{alltt}


Theorem decomp :
  {\prodsym} (A : Set) (n : nat) (v : vector A (S n)),
  v = Vcons (Vhead v) (Vtail v).
Proof.
 intros.
 change (Vcons (Vhead v) (Vtail v)) with (Vid _  (S n) v).
\it
 1 subgoal
  
  A : Set
  n : nat
  v : vector A (S n)
  ============================
   v = Vid A (S n) v

\tt{} apply Vid_eq.
Defined.
\end{alltt}


Both lemmas: \citecoq{zero\_nil} and \citecoq{decomp},
can be used to easily derive a double recursion principle
on vectors of same length:


\begin{alltt}
Definition vector_double_rect : 
    {\prodsym} (A:Set) (P: {\prodsym} (n:nat),(vector A n){\arrow}(vector A n) {\arrow} Type),
        P 0 Vnil Vnil {\arrow}
        ({\prodsym} n (v1 v2 : vector A n) a b, P n v1 v2 {\arrow}
             P (S n) (Vcons a v1) (Vcons  b v2)) {\arrow}
        {\prodsym} n (v1 v2 : vector A n), P n v1 v2.
 induction n.
 intros; rewrite (zero_nil _ v1); rewrite (zero_nil _ v2).
 auto.
 intros v1 v2; rewrite (decomp _ _ v1);rewrite (decomp _ _ v2).
 apply X0; auto.
Defined.
\end{alltt}

Notice that, due to the conversion rule of {\coq}'s type system,
this function can be used directly with \citecoq{Prop} or \citecoq{Set}
instead of type (thus it is useless to build 
\citecoq{vector\_double\_ind} and \citecoq{vector\_double\_rec}) from scratch.

We finish this example with showing how to define the bitwise
\emph{or} on boolean vectors of the same length, 
and proving a little property about this
operation.

\begin{alltt}
Definition bitwise_or n v1 v2 : vector bool n :=
   vector_double_rect 
    bool 
    (fun n v1 v2 {\funarrow} vector bool n)
    Vnil
    (fun n v1 v2 a b r {\funarrow} Vcons (orb a b) r) n v1 v2.
\end{alltt}

Let us  define recursively the $n$-th element of a vector. Notice
that it must be a partial function, in case $n$ is greater or equal
than the length of the vector. Since {\coq} only considers total
functions, the function returns a value in an \emph{option} type.

\begin{alltt}
Fixpoint vector_nth (A:Set)(n:nat)(p:nat)(v:vector A p)
                  \{struct v\}
                  : option A :=
  match n,v  with
    _   , Vnil {\funarrow} None
  | 0   , Vcons b  _ _ {\funarrow} Some b
  | S n', Vcons _  p' v' {\funarrow} vector_nth A n'  p' v'
  end.
Implicit Arguments vector_nth [A p].
\end{alltt}

We can now prove --- using the double induction combinator ---
a simple property relying \citecoq{vector\_nth} and \citecoq{bitwise\_or}:

\begin{alltt}
Lemma nth_bitwise :
   {\prodsym} (n:nat) (v1 v2: vector bool n) i  a b,
      vector_nth i v1 = Some a {\arrow}
      vector_nth i v2 = Some b {\arrow}
      vector_nth i (bitwise_or _ v1 v2) = Some (orb a b).
Proof.
 intros  n v1 v2; pattern n,v1,v2.
 apply vector_double_rect.
 simpl.
 destruct i; discriminate 1.
 destruct i; simpl;auto.
 injection 1; injection 2;intros; subst a; subst b; auto.
Qed.
\end{alltt}


\section{Co-inductive Types and Non-ending Constructions}
\label{CoInduction}

The objects of an inductive type are well-founded with respect to
the constructors of the type. In other words, these objects are built
by applying \emph{a finite number of times} the constructors of the type.
Co-inductive types are obtained by relaxing this condition,
and may contain non-well-founded objects \cite{EG96,EG95a}.  An
example of a co-inductive type is the type of  infinite
sequences formed with elements of type $A$, also called streams.  This
type can be introduced through the following definition:

\begin{alltt}
 CoInductive Stream (A: Set) :Set   := 
 | Cons : A\arrow{}Stream A\arrow{}Stream A.
\end{alltt}

If we are interested in finite or infinite sequences, we consider the type
of \emph{lazy lists}:

\begin{alltt}
CoInductive LList (A: Set) : Set :=
 |  LNil : LList A
 |  LCons : A {\arrow} LList A {\arrow} LList A.
\end{alltt}


It is also possible to define  co-inductive types for the 
trees with infinite branches (see Chapter 13 of~\cite{coqart}).

Structural induction is the way of expressing that inductive types
only contain well-founded objects. Hence, this elimination principle
is not valid for co-inductive types, and the only elimination rule for
streams is case analysis.  This principle can be used, for example, to
define the destructors \textsl{head} and \textsl{tail}.

\begin{alltt}
 Definition head (A:Set)(s : Stream A) := 
   match s with Cons a s' {\funarrow} a end.

 Definition tail (A : Set)(s : Stream A) :=
      match s with Cons a s' {\funarrow} s' end.
\end{alltt}

Infinite objects are defined by means of (non-ending) methods of
construction, like in lazy functional programming languages.  Such
methods can be defined using the \texttt{CoFixpoint} command
\refmancite{Section \ref{CoFixpoint}}. For example, the following
definition introduces the infinite list $[a,a,a,\ldots]$:

\begin{alltt}
 CoFixpoint repeat (A:Set)(a:A) : Stream A := 
   Cons a (repeat a).
\end{alltt}


However, not every co-recursive definition is an admissible method of
construction. Similarly to the case of structural induction, the
definition must verify a \textsl{guardedness} condition to be
accepted. This condition states that any recursive call in the
definition must be protected --i.e, be an argument of-- some
constructor, and only an argument of constructors \cite{EG94a}. The
following definitions are examples of valid methods of construction:

\begin{alltt}
CoFixpoint iterate (A: Set)(f: A {\arrow} A)(a : A) : Stream A:=
    Cons a (iterate f (f a)).

CoFixpoint map 
  (A B:Set)(f: A {\arrow} B)(s : Stream A) : Stream B:=
  match s with Cons a tl {\funarrow} Cons (f a) (map f tl) end.
\end{alltt}

\begin{exercise}
Define two different methods for constructing the stream which 
infinitely alternates the values \citecoq{true} and \citecoq{false}.
\end{exercise}
\begin{exercise}
Using the destructors \texttt{head} and \texttt{tail}, define a function
which takes the n-th element of an infinite stream.
\end{exercise}

A non-ending method of construction is computed lazily. This means
that its definition is unfolded only when the object that it
introduces is eliminated, that is, when it appears as the argument of
a case expression. We can check this using the command
\texttt{Eval}.

\begin{alltt}
Eval simpl in (fun (A:Set)(a:A) {\funarrow} repeat a).
\it  = fun (A : Set) (a : A) {\funarrow} repeat a
     : {\prodsym} A : Set, A {\arrow} Stream A
\tt
Eval simpl in (fun (A:Set)(a:A) {\funarrow} head (repeat a)).
\it  = fun (A : Set) (a : A) {\funarrow} a
     : {\prodsym} A : Set, A {\arrow} A
\end{alltt}

%\begin{exercise}
%Prove the following theorem:
%\begin{verbatim}
%Theorem expand_repeat : (a:A)(repeat a)=(Cons a (repeat a)). 
%\end{verbatim}
%Hint: Prove first the streams version of the lemma in exercise 
%\ref{expand}.
%\end{exercise}

\subsection{Extensional Properties}

Case analysis is also a valid proof principle for infinite
objects. However, this principle is not sufficient to prove
\textsl{extensional} properties, that is, properties concerning the
whole infinite object \cite{EG95a}. A typical example of an
extensional property is the predicate expressing that two streams have
the same elements. In many cases, the minimal reflexive relation $a=b$
that is used as equality for inductive types is too small to capture
equality between streams. Consider for example the streams
$\texttt{iterate}\;f\;(f\;x)$ and
$(\texttt{map}\;f\;(\texttt{iterate}\;f\;x))$. Even though these two streams have
the same elements, no finite expansion of their definitions lead to
equal terms. In other words, in order to deal with extensional
properties, it is necessary to construct infinite proofs. The type of
infinite proofs of equality can be introduced as a co-inductive
predicate, as follows:
\begin{alltt}
CoInductive EqSt (A: Set) : Stream A {\arrow} Stream A {\arrow} Prop :=
  eqst : {\prodsym} s1 s2: Stream A,
      head s1 = head s2 {\arrow}
      EqSt (tail s1) (tail s2) {\arrow}
      EqSt s1 s2.
\end{alltt}

It is possible to introduce proof principles for reasoning about
infinite objects as combinators defined through
\texttt{CoFixpoint}. However, oppositely to the case of inductive
types, proof principles associated to co-inductive types are not
elimination but \textsl{introduction} combinators. An example of such
a combinator is Park's principle for proving the equality of two
streams, usually called the \textsl{principle of co-induction}. It
states that two streams are equal if they satisfy a
\textit{bisimulation}.  A bisimulation is a binary relation $R$ such
that any pair of streams $s_1$ ad $s_2$ satisfying $R$ have equal
heads, and tails also satisfying $R$.  This principle is in fact a
method for constructing an infinite proof:

\begin{alltt}
Section Parks_Principle.
Variable A : Set.
Variable    R      : Stream A {\arrow} Stream A {\arrow} Prop.
Hypothesis  bisim1 : {\prodsym} s1 s2:Stream A, 
                       R s1 s2 {\arrow} head s1 = head s2.

Hypothesis  bisim2 : {\prodsym} s1 s2:Stream A, 
                       R s1 s2 {\arrow} R (tail s1) (tail s2).

CoFixpoint park_ppl     : 
 {\prodsym} s1 s2:Stream A, R s1 s2 {\arrow} EqSt s1 s2 :=
 fun s1 s2 (p : R s1 s2) {\funarrow}
      eqst s1 s2 (bisim1 s1 s2 p) 
                 (park_ppl (tail s1) 
                           (tail s2) 
                           (bisim2 s1 s2 p)).
End Parks_Principle.
\end{alltt}

Let us use the principle of co-induction to prove the extensional
equality mentioned above. 
\begin{alltt}
Theorem map_iterate : {\prodsym} (a:Set)(f:A{\arrow}A)(x:A),
                       EqSt (iterate f (f x)) 
                            (map f (iterate f x)).
Proof.
 intros A f x.
 apply park_ppl with
   (R:= fun s1 s2 {\funarrow}
       {\exsym} x: A, s1 = iterate f (f x) {\coqand} 
                s2 = map f (iterate f x)).

 intros s1 s2 (x0,(eqs1,eqs2));
    rewrite eqs1; rewrite eqs2; reflexivity.
 intros s1 s2 (x0,(eqs1,eqs2)).
 exists (f x0);split;
    [rewrite eqs1|rewrite eqs2]; reflexivity.
 exists x;split; reflexivity.
Qed.
\end{alltt}

The use of Park's principle is sometimes annoying, because it requires
to find an invariant relation and prove that it is indeed a
bisimulation.  In many cases, a shorter proof can be obtained trying
to construct an ad-hoc infinite proof, defined by a guarded
declaration.  The tactic ``~``\texttt{Cofix $f$}~'' can be used to do
that. Similarly to the tactic \texttt{fix} indicated in Section
\ref{WellFoundedRecursion}, this tactic introduces an extra hypothesis
$f$ into the context, whose type is the same as the current goal. Note
that the applications of $f$ in the proof \textsl{must be guarded}. In
order to prevent us from doing unguarded calls, we can define a tactic
that always apply a constructor before using $f$ \refmancite{Chapter
\ref{WritingTactics}} :

\begin{alltt}
Ltac infiniteproof f :=
  cofix f; 
  constructor; 
  [clear f| simpl; try (apply f; clear f)].
\end{alltt}


In the example above, this tactic produces a much simpler proof
that the former one:

\begin{alltt}
Theorem map_iterate' : {\prodsym} ((A:Set)f:A{\arrow}A)(x:A),
                       EqSt (iterate f (f x))
                            (map f (iterate f x)).
Proof.
 infiniteproof map_iterate'.
 reflexivity.
Qed.
\end{alltt}

\begin{exercise}
Define a co-inductive type $Nat$ containing non-standard 
natural numbers --this is, verifying 

$$\exists m  \in \mbox{\texttt{Nat}}, \forall\, n \in \mbox{\texttt{Nat}}, n<m$$.
\end{exercise}

\begin{exercise}
Prove that the extensional equality of streams is an equivalence relation
using Park's co-induction principle.
\end{exercise}


\begin{exercise}
Provide a suitable definition of ``being an ordered list'' for infinite lists
and define a principle for proving that an infinite list is ordered. Apply
this method to the list $[0,1,\ldots ]$. Compare the result with 
exercise \ref{zeroton}.
\end{exercise}

\subsection{About injection, discriminate, and inversion}
Since co-inductive types are closed w.r.t. their constructors,
the techniques shown in Section~\ref{CaseTechniques} work also 
with these types.

Let us consider the type of lazy lists, introduced on page~\pageref{CoInduction}.
The following lemmas are straightforward applications
 of \texttt{discriminate} and \citecoq{injection}:

\begin{alltt}
Lemma Lnil_not_Lcons : {\prodsym} (A:Set)(a:A)(l:LList A),
                               LNil {\coqdiff} (LCons a l).
Proof.
 intros;discriminate.
Qed.

Lemma injection_demo : {\prodsym} (A:Set)(a b : A)(l l': LList A),
                       LCons a (LCons b l) = LCons b (LCons a l') {\arrow}
                       a = b {\coqand} l = l'.
Proof.
 intros A a b l l' e; injection e; auto.
Qed.

\end{alltt}

In order to show \citecoq{inversion} at work, let us define
two predicates on lazy lists:

\begin{alltt}
Inductive Finite (A:Set) : LList A {\arrow} Prop :=
|  Lnil_fin : Finite (LNil (A:=A))
|  Lcons_fin : {\prodsym} a l, Finite l {\arrow} Finite (LCons a l).

CoInductive Infinite  (A:Set) : LList A {\arrow} Prop :=
| LCons_inf : {\prodsym} a l, Infinite l {\arrow} Infinite (LCons a l).
\end{alltt}

\noindent
First, two easy theorems:
\begin{alltt}
Lemma LNil_not_Infinite : {\prodsym} (A:Set), ~ Infinite (LNil (A:=A)).
Proof.
  intros A H;inversion H.
Qed.

Lemma Finite_not_Infinite : {\prodsym} (A:Set)(l:LList A),
                                Finite l {\arrow} ~ Infinite l.
Proof.
 intros A l H; elim H.
 apply LNil_not_Infinite.
 intros a l0 F0 I0' I1.
 case I0'; inversion_clear I1.
 trivial.
Qed.
\end{alltt}


On the other hand, the next proof uses the \citecoq{cofix} tactic. 
Notice the destructuration of \citecoq{l}, which allows us to
apply  the constructor \texttt{LCons\_inf}, thus  satisfying
 the guard condition: 
\begin{alltt}
Lemma Not_Finite_Infinite : {\prodsym} (A:Set)(l:LList A),
                            ~ Finite l {\arrow} Infinite l.
Proof.
 cofix H.
 destruct l.
 intro; 
  absurd (Finite (LNil (A:=A)));
  [auto|constructor].
\it




1 subgoal
  
  H : forall (A : Set) (l : LList A), ~ Finite l -> Infinite l
  A : Set
  a : A
  l : LList A
  H0 : ~ Finite (LCons a l)
  ============================
   Infinite l
\end{alltt}
At this point, one must not apply \citecoq{H}! . It would be possible
to solve the current goal by an inversion of ``~\citecoq{Finite (LCons a l)}~'', but, since the guard condition would be violated, the user
would get an error message after typing \citecoq{Qed}.
In order to satisfy the guard condition, we apply the constructor of
\citecoq{Infinite}, \emph{then} apply \citecoq{H}.

\begin{alltt}
 constructor.
 apply H.
 red; intro H1;case H0.
 constructor.
 trivial.
Qed.
\end{alltt}




The reader is invited to replay this proof and understand each of its steps.


\bibliographystyle{abbrv}
\bibliography{manbiblio,morebib}

\end{document}