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diff --git a/test-suite/success/decl_mode.v b/test-suite/success/decl_mode.v deleted file mode 100644 index 58f79d45..00000000 --- a/test-suite/success/decl_mode.v +++ /dev/null @@ -1,182 +0,0 @@ -(* \sqrt 2 is irrationnal, (c) 2006 Pierre Corbineau *) - -Set Firstorder Depth 1. -Require Import ArithRing Wf_nat Peano_dec Div2 Even Lt. - -Lemma double_div2: forall n, div2 (double n) = n. -proof. - assume n:nat. - per induction on n. - suppose it is 0. - suffices (0=0) to show thesis. - thus thesis. - suppose it is (S m) and Hrec:thesis for m. - have (div2 (double (S m))= div2 (S (S (double m)))). - ~= (S (div2 (double m))). - thus ~= (S m) by Hrec. - end induction. -end proof. -Show Script. -Qed. - -Lemma double_inv : forall n m, double n = double m -> n = m . -proof. - let n, m be such that H:(double n = double m). -have (n = div2 (double n)) by double_div2,H. - ~= (div2 (double m)) by H. - thus ~= m by double_div2. -end proof. -Qed. - -Lemma double_mult_l : forall n m, (double (n * m)=n * double m). -proof. - assume n:nat and m:nat. - have (double (n * m) = n*m + n * m). - ~= (n * (m+m)) by * using ring. - thus ~= (n * double m). -end proof. -Qed. - -Lemma double_mult_r : forall n m, (double (n * m)=double n * m). -proof. - assume n:nat and m:nat. - have (double (n * m) = n*m + n * m). - ~= ((n + n) * m) by * using ring. - thus ~= (double n * m). -end proof. -Qed. - -Lemma even_is_even_times_even: forall n, even (n*n) -> even n. -proof. - let n be such that H:(even (n*n)). - per cases of (even n \/ odd n) by even_or_odd. - suppose (odd n). - hence thesis by H,even_mult_inv_r. - end cases. -end proof. -Qed. - -Lemma main_thm_aux: forall n,even n -> -double (double (div2 n *div2 n))=n*n. -proof. - given n such that H:(even n). - *** have (double (double (div2 n * div2 n)) - = double (div2 n) * double (div2 n)) - by double_mult_l,double_mult_r. - thus ~= (n*n) by H,even_double. -end proof. -Qed. - -Require Import Omega. - -Lemma even_double_n: (forall m, even (double m)). -proof. - assume m:nat. - per induction on m. - suppose it is 0. - thus thesis. - suppose it is (S mm) and thesis for mm. - then H:(even (S (S (mm+mm)))). - have (S (S (mm + mm)) = S mm + S mm) using omega. - hence (even (S mm +S mm)) by H. - end induction. -end proof. -Qed. - -Theorem main_theorem: forall n p, n*n=double (p*p) -> p=0. -proof. - assume n0:nat. - define P n as (forall p, n*n=double (p*p) -> p=0). - claim rec_step: (forall n, (forall m,m<n-> P m) -> P n). - let n be such that H:(forall m : nat, m < n -> P m) and p:nat . - per cases of ({n=0}+{n<>0}) by eq_nat_dec. - suppose H1:(n=0). - per cases on p. - suppose it is (S p'). - assume (n * n = double (S p' * S p')). - =~ 0 by H1,mult_n_O. - ~= (S ( p' + p' * S p' + S p'* S p')) - by plus_n_Sm. - hence thesis . - suppose it is 0. - thus thesis. - end cases. - suppose H1:(n<>0). - assume H0:(n*n=double (p*p)). - have (even (double (p*p))) by even_double_n . - then (even (n*n)) by H0. - then H2:(even n) by even_is_even_times_even. - then (double (double (div2 n *div2 n))=n*n) - by main_thm_aux. - ~= (double (p*p)) by H0. - then H':(double (div2 n *div2 n)= p*p) by double_inv. - have (even (double (div2 n *div2 n))) by even_double_n. - then (even (p*p)) by even_double_n,H'. - then H3:(even p) by even_is_even_times_even. - have (double(double (div2 n * div2 n)) = n*n) - by H2,main_thm_aux. - ~= (double (p*p)) by H0. - ~= (double(double (double (div2 p * div2 p)))) - by H3,main_thm_aux. - then H'':(div2 n * div2 n = double (div2 p * div2 p)) - by double_inv. - then (div2 n < n) by lt_div2,neq_O_lt,H1. - then H4:(div2 p=0) by (H (div2 n)),H''. - then (double (div2 p) = double 0). - =~ p by even_double,H3. - thus ~= 0. - end cases. - end claim. - hence thesis by (lt_wf_ind n0 P). -end proof. -Qed. - -Require Import Reals Field. -(*Coercion INR: nat >->R. -Coercion IZR: Z >->R.*) - -Open Scope R_scope. - -Lemma square_abs_square: - forall p,(INR (Z.abs_nat p) * INR (Z.abs_nat p)) = (IZR p * IZR p). -proof. - assume p:Z. - per cases on p. - suppose it is (0%Z). - thus thesis. - suppose it is (Zpos z). - thus thesis. - suppose it is (Zneg z). - have ((INR (Z.abs_nat (Zneg z)) * INR (Z.abs_nat (Zneg z))) = - (IZR (Zpos z) * IZR (Zpos z))). - ~= ((- IZR (Zpos z)) * (- IZR (Zpos z))). - thus ~= (IZR (Zneg z) * IZR (Zneg z)). - end cases. -end proof. -Qed. - -Definition irrational (x:R):Prop := - forall (p:Z) (q:nat),q<>0%nat -> x<> (IZR p/INR q). - -Theorem irrationnal_sqrt_2: irrational (sqrt (INR 2%nat)). -proof. - let p:Z,q:nat be such that H:(q<>0%nat) - and H0:(sqrt (INR 2%nat)=(IZR p/INR q)). - have H_in_R:(INR q<>0:>R) by H. - have triv:((IZR p/INR q* INR q) =IZR p :>R) by * using field. - have sqrt2:((sqrt (INR 2%nat) * sqrt (INR 2%nat))= INR 2%nat:>R) by sqrt_def. - have (INR (Z.abs_nat p * Z.abs_nat p) - = (INR (Z.abs_nat p) * INR (Z.abs_nat p))) - by mult_INR. - ~= (IZR p* IZR p) by square_abs_square. - ~= ((IZR p/INR q*INR q)*(IZR p/INR q*INR q)) by triv. (* we have to factor because field is too weak *) - ~= ((IZR p/INR q)*(IZR p/INR q)*(INR q*INR q)) using ring. - ~= (sqrt (INR 2%nat) * sqrt (INR 2%nat)*(INR q*INR q)) by H0. - ~= (INR (2%nat * (q*q))) by sqrt2,mult_INR. - then (Z.abs_nat p * Z.abs_nat p = 2* (q * q))%nat. - ~= ((q*q)+(q*q))%nat. - ~= (Div2.double (q*q)). - then (q=0%nat) by main_theorem. - hence thesis by H. -end proof. -Qed. |