5 files changed, 337 insertions, 1 deletions

diff --git a/test-suite/success/Typeclasses.v b/test-suite/success/Typeclasses.v
index 30a2a7429..d6e590af3 100644
--- a/test-suite/success/Typeclasses.v
+++ b/test-suite/success/Typeclasses.v
@@ -57,4 +57,25 @@ Section sec.
  let's try to get rid of the intermediate constant foo.
  Surely we can just expand it inline, right? Wrong!: *)
  Check U (fun x => e x) _.
-End sec.
\ No newline at end of file
+End sec.
+
+Module IterativeDeepening.
+
+  Class A.
+  Class B.
+  Class C.
+
+  Instance: B -> A | 0.
+  Instance: C -> A | 0.
+  Instance: C -> B -> A | 0.
+  Instance: A -> A | 0.
+  
+  Goal C -> A.
+    intros.
+    Set Typeclasses Debug.
+    Fail Timeout 1 typeclasses eauto.
+    Set Typeclasses Iterative Deepening.
+    typeclasses eauto.
+  Qed.
+
+End IterativeDeepening.
diff --git a/test-suite/success/decl_mode2.v b/test-suite/success/decl_mode2.v
new file mode 100644
index 000000000..46174e481
--- /dev/null
+++ b/test-suite/success/decl_mode2.v
@@ -0,0 +1,249 @@
+Theorem this_is_trivial: True.
+proof.
+  thus thesis.
+end proof.
+Qed.
+
+Theorem T: (True /\ True) /\ True.
+  split. split.
+proof. (* first subgoal *)
+  thus thesis.
+end proof.
+trivial. (* second subgoal *)
+proof. (* third subgoal *)
+  thus thesis.
+end proof.
+Abort.
+
+Theorem this_is_not_so_trivial: False.
+proof.
+end proof. (* here a warning is issued *)
+Fail Qed. (* fails: the proof in incomplete *)
+Admitted. (* Oops! *)
+
+Theorem T: True.
+proof.
+escape.
+auto.
+return.
+Abort.
+
+Theorem T: let a:=false in let b:= true in ( if a then True else False -> if b then True else False).
+intros a b.
+proof.
+assume H:(if a then True else False).
+reconsider H as False.
+reconsider thesis as True.
+Abort.
+
+Theorem T: forall x, x=2 -> 2+x=4.
+proof.
+let x be such that H:(x=2).
+have H':(2+x=2+2) by H.
+Abort.
+
+Theorem T: forall x, x=2 -> 2+x=4.
+proof.
+let x be such that H:(x=2).
+then (2+x=2+2).
+Abort.
+
+Theorem T: forall x, x=2 -> x + x = x * x.
+proof.
+let x be such that H:(x=2).
+have (4 = 4).
+        ~= (2 * 2).
+        ~= (x * x) by H.
+        =~ (2 + 2).
+        =~ H':(x + x) by H.
+Abort.
+
+Theorem T: forall x, x + x = x * x -> x = 0 \/ x = 2.
+proof.
+let x be such that H:(x + x = x * x).
+claim H':((x - 2) * x = 0).
+thus thesis. 
+end claim.
+Abort.
+
+Theorem T: forall (A B:Prop), A -> B -> A /\ B.
+intros A B HA HB.
+proof.
+hence B.
+Abort.
+
+Theorem T: forall (A B C:Prop), A -> B -> C -> A /\ B /\ C.
+intros A B C HA HB HC.
+proof.
+thus B by HB.
+Abort.
+
+Theorem T: forall (A B C:Prop), A -> B -> C -> A /\ B.
+intros A B C HA HB HC.
+proof.
+Fail hence C. (* fails *)
+Abort.
+
+Theorem T: forall (A B:Prop), B -> A \/ B.
+intros A B HB.
+proof.
+hence B.
+Abort.
+
+Theorem T: forall (A B C D:Prop), C -> D -> (A /\ B) \/ (C /\ D).
+intros A B C D HC HD.
+proof.
+thus C by HC.
+Abort.
+
+Theorem T: forall (P:nat -> Prop), P 2 -> exists x,P x.
+intros P HP.
+proof.
+take 2.
+Abort.
+
+Theorem T: forall (P:nat -> Prop), P 2 -> exists x,P x.
+intros P HP.
+proof.
+hence (P 2).
+Abort.
+
+Theorem T: forall (P:nat -> Prop) (R:nat -> nat -> Prop), P 2 -> R 0 2 -> exists x, exists y, P y /\ R x y.
+intros P R HP HR.
+proof.
+thus (P 2) by HP.
+Abort.
+
+Theorem T: forall (A B:Prop) (P:nat -> Prop), (forall x, P x -> B) -> A -> A /\ B.
+intros A B P HP HA.
+proof.
+suffices to have x such that HP':(P x) to show B by HP,HP'.
+Abort.
+
+Theorem T: forall (A:Prop) (P:nat -> Prop), P 2 -> A -> A /\ (forall x, x = 2 -> P x).
+intros A P HP HA.
+proof.
+(* BUG: the next line fails when it should succeed.
+Waiting for someone to investigate the bug.
+focus on (forall x, x = 2 -> P x).
+let x be such that (x = 2).
+hence thesis by HP.
+end focus.
+*)
+Abort.
+
+Theorem T: forall x, x = 0 -> x + x = x * x.
+proof.
+let x be such that H:(x = 0).
+define sqr x as (x * x).
+reconsider thesis as (x + x = sqr x).
+Abort.
+
+Theorem T: forall (P:nat -> Prop), forall x, P x -> P x.
+proof.
+let P:(nat -> Prop).
+let x:nat.
+assume HP:(P x).
+Abort.
+
+Theorem T: forall (P:nat -> Prop), forall x, P x -> P x.
+proof.
+let P:(nat -> Prop).
+Fail let x. (* fails because x's type is not clear *) 
+let x be such that HP:(P x). (* here x's type is inferred from (P x) *)
+Abort.
+
+Theorem T: forall (P:nat -> Prop), forall x, P x -> P x -> P x.
+proof.
+let P:(nat -> Prop).
+let x:nat.
+assume (P x). (* temporary name created *)
+Abort.
+
+Theorem T: forall (P:nat -> Prop), forall x, P x -> P x.
+proof.
+let P:(nat -> Prop).
+let x be such that (P x). (* temporary name created *)
+Abort.
+
+Theorem T: forall (P:nat -> Prop) (A:Prop), (exists x, (P x /\ A)) -> A.
+proof.
+let P:(nat -> Prop),A:Prop be such that H:(exists x, P x /\ A).
+consider x such that HP:(P x) and HA:A from H.
+Abort.
+
+(* Here is an example with pairs: *)
+
+Theorem T: forall p:(nat * nat)%type, (fst p >= snd p) \/ (fst p < snd p).
+proof.
+let p:(nat * nat)%type.
+consider x:nat,y:nat from p.
+reconsider thesis as (x >= y \/ x < y). 
+Abort.
+
+Theorem T: forall P:(nat -> Prop), (forall n, P n -> P (n - 1)) -> 
+(exists m, P m) -> P 0.
+proof.
+let P:(nat -> Prop) be such that HP:(forall n, P n -> P (n - 1)).
+given m such that Hm:(P m).  
+Abort.
+
+Theorem T: forall (A B C:Prop), (A -> C) -> (B -> C) -> (A \/ B) -> C.
+proof.
+let A:Prop,B:Prop,C:Prop be such that HAC:(A -> C) and HBC:(B -> C).
+assume HAB:(A \/ B).
+per cases on HAB.
+suppose A.
+  hence thesis by HAC.
+suppose HB:B.
+  thus thesis by HB,HBC.
+end cases.
+Abort.
+
+Section Coq.
+
+Hypothesis EM : forall P:Prop, P \/ ~ P.
+
+Theorem T: forall (A C:Prop), (A -> C) -> (~A -> C) -> C.
+proof.
+let A:Prop,C:Prop be such that HAC:(A -> C) and HNAC:(~A -> C).
+per cases of (A \/ ~A) by EM.
+suppose (~A).
+  hence thesis by HNAC.
+suppose A.
+  hence thesis by HAC.
+end cases.
+Abort.
+
+Theorem T: forall (A C:Prop), (A -> C) -> (~A -> C) -> C.
+proof.
+let A:Prop,C:Prop be such that HAC:(A -> C) and HNAC:(~A -> C).
+per cases on (EM A).
+suppose (~A).
+Abort.
+End Coq.
+
+Theorem T: forall (A B:Prop) (x:bool), (if x then A else B) -> A \/ B.
+proof.
+let A:Prop,B:Prop,x:bool.
+per cases on x.
+suppose it is true.
+  assume A.
+  hence A.
+suppose it is false.
+  assume B.
+  hence B.
+end cases.
+Abort.
+
+Theorem T: forall (n:nat), n + 0 = n.
+proof.
+let n:nat.
+per induction on n.
+suppose it is 0.
+  thus (0 + 0 = 0).
+suppose it is (S m) and H:thesis for m.
+  then (S (m + 0) = S m).
+  thus =~ (S m + 0).
+end induction.
+Abort.
\ No newline at end of file
diff --git a/test-suite/success/shrink_abstract.v b/test-suite/success/shrink_abstract.v
new file mode 100644
index 000000000..3f6b9cb39
--- /dev/null
+++ b/test-suite/success/shrink_abstract.v
@@ -0,0 +1,13 @@
+Set Shrink Abstract.
+
+Definition foo : forall (n m : nat), bool.
+Proof.
+pose (p := 0).
+intros n.
+pose (q := n).
+intros m.
+pose (r := m).
+abstract (destruct m; [left|right]).
+Defined.
+
+Check (foo_subproof : nat -> bool).
diff --git a/test-suite/success/shrink_obligations.v b/test-suite/success/shrink_obligations.v
new file mode 100644
index 000000000..676b97878
--- /dev/null
+++ b/test-suite/success/shrink_obligations.v
@@ -0,0 +1,28 @@
+Require Program.
+
+Obligation Tactic := idtac.
+
+Set Shrink Obligations.
+
+Program Definition foo (m : nat) (p := S m) (n : nat) (q := S n) : unit :=
+let bar : {r | n < r} := _ in
+let qux : {r | p < r} := _ in
+let quz : m = n -> True := _ in
+tt.
+Next Obligation.
+intros m p n q.
+exists (S n); constructor.
+Qed.
+Next Obligation.
+intros m p n q.
+exists (S (S m)); constructor.
+Qed.
+Next Obligation.
+intros m p n q ? ? H.
+destruct H.
+constructor.
+Qed.
+
+Check (foo_obligation_1 : forall n, {r | n < r}).
+Check (foo_obligation_2 : forall m, {r | (S m) < r}).
+Check (foo_obligation_3 : forall m n, m = n -> True).
diff --git a/test-suite/success/subst.v b/test-suite/success/subst.v
new file mode 100644
index 000000000..8336f6a80
--- /dev/null
+++ b/test-suite/success/subst.v
@@ -0,0 +1,25 @@
+(* Test various subtleties of the "subst" tactics *)
+
+(* Should proceed from left to right (see #4222) *)
+Goal forall x y, x = y -> x = 3 -> y = 2 -> x = y.
+intros.
+subst.
+change (3 = 2) in H1.
+change (3 = 3).
+Abort.
+
+(* Should work with "x = y" and "x = t" equations (see #4214, failed in 8.4) *)
+Goal forall x y, x = y -> x = 3 -> x = y.
+intros.
+subst.
+change (3 = 3).
+Abort.
+
+(* Should substitute cycles once, until a recursive equation is obtained *)
+(* (failed in 8.4) *)
+Goal forall x y, x = S y -> y = S x -> x = y.
+intros.
+subst.
+change (y = S (S y)) in H0.
+change (S y = y).
+Abort.