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Diffstat (limited to 'test-suite/success/decl_mode2.v')
| -rw-r--r-- | test-suite/success/decl_mode2.v | 249 |
1 files changed, 0 insertions, 249 deletions
diff --git a/test-suite/success/decl_mode2.v b/test-suite/success/decl_mode2.v deleted file mode 100644 index 46174e48..00000000 --- a/test-suite/success/decl_mode2.v +++ /dev/null @@ -1,249 +0,0 @@ -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.
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