Coq将不存在转换为forall语句 [英] Coq convert non exist to forall statement

问题描述

我是Coq的新手。这是我的问题。
我有一条声明说：

I'm new to Coq. Here's my problem. I have a statement says:

H : forall x : term, ~ (exists y : term, P x y /\ ~ P y x)

我想它等同于：

forall x y : term, (P x y /\ ~ P y x) -> false

但是我可以使用哪种策略转换假设？

But which tactic can I use to convert the hypothesis?

推荐答案

我不知道将不存在转换为永久不存在的策略，但是您总是可以 $ c>并证明这一点。 （如果您反复需要，则可以将其打包到和 Ltac 策略定义或简单的定理[1]。）

I don't know of a tactic to turn not-exists into forall-not, but you can always just assert and prove it. (If you need that repeatedly, you can pack that up into an Ltac tactic definition or a simple theorem[1].)

以下是证明这一点的三种方法。 （您应该能够将此成绩单复制/粘贴到CoqIDE或Emacs / ProofGeneral中并逐步执行代码。）

Here's three ways of getting this proved. (You should be able to just copy/paste this transcript into CoqIDE or Emacs/ProofGeneral and step through the code.)

[1]存在引理<$ c库中的$ c> not_ex_all_not Coq.Logic.Classical_Pred_Type ，但是加载会引入经典逻辑的公理（这里甚至不需要）。

[1] There exists a lemma not_ex_all_not in the Library Coq.Logic.Classical_Pred_Type, but loading that would pull in the axiom for classical logic (which isn't even needed here).

(* dummy context to set up H of the correct type, for testing it out *)
Lemma SomeName (term : Type) (P : term -> term -> Prop) :
  (forall x : term, ~ (exists (y : term), P x y /\ ~ P y x)) ->
  True. (* dummy goal *)
Proof.
  intro H.
  (* end dummy context *)

（* 这是长版本，有一些解释： *）

(* Here's the long version, with some explanations: *)

  (* this states the goal, result will be put into the context as H' *)
  assert (forall (x y : term), (P x y /\ ~ P y x) -> False) as H'.
    (* get rid of unneeded variables in context, get new args *)
    clear - H; intros x y Pxy.
    (* unfolding the not shows the computational structure of this *)
    unfold not at 1 in H.
    (* yay... (x, y, Pxy) will produce False via H *)
    (* specialize to x and apply... *)
    apply (H x).
    (* ...and provide the witness. *)
    exists y.  exact Pxy.
    (* done. *)

  (* let's do it again... *)
  clear H'.

（* 您也可以在一条语句： *）

(* you can also do it in a single statement: *)

  assert (forall x y, (P x y /\ ~ P y x) -> False) as H'
    by (clear -H; intros x y Pxy; apply (H x (ex_intro _ y Pxy))).

  (* and again... *)
  clear H'.

（* 这样简单的东西也可以可以用手写来写： *）

(* simple stuff like this can also be written by hand: *)

  pose proof (fun x y Pxy => H x (ex_intro _ y Pxy)) as H'; simpl in H'.

（* 现在您有H'正确的类型；可以选择删除旧的H： *）

(* now you have H' of the right type; optionally get rid of the old H: *)

  clear H; rename H' into H.

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