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)
I guess it is equivalent to:
forall x y : term, (P x y /\ ~ P y x) -> false
But which tactic can I use to convert the hypothesis?
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].)
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] 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 *)
(*
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'.
(*
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'.
(*
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'.
(*
now you have H' of the right type; optionally get rid of the old H: *)
clear H; rename H' into H.
You can use unfold not at 1 in H
. ~ P
is just notation for not P
, and not P = (P -> False)
by definition. The at 1
part means you only want to unfold
the first occurrence of not
, and the in H
part means you only want to unfold
it in hypothesis H
.
来源:https://stackoverflow.com/questions/21644359/coq-convert-non-exist-to-forall-statement