7

I'm trying to prove a substitution theorem about Prop, and I'm failing miserably. Can the following theorem be proven in coq, and if not, why not.

  Theorem prop_subst:
    forall (f : Prop -> Prop) (P Q : Prop), 
      (P <-> Q) -> ((f P) <-> (f Q)).

The point is that the proof, in logic, would be by induction. Prop isn't defined inductively, as far as I can see. How would such a theorem be proven in Coq?

Volker Stolz
  • 7,274
  • 1
  • 32
  • 50
Mayer Goldberg
  • 1,378
  • 11
  • 23

3 Answers3

8

Here's the answer: The property I was looking for is called propositional extensionality, and means that forall p q : Prop, (p <-> q) -> (p = q). The converse, is trivial. This is something that is defined in Library Coq.Logic.ClassicalFacts, together with other facts from classical, i.e., non-intuitionistic logic. The above definition is called prop_extensionality, and can be used as follows: Axiom EquivThenEqual: prop_extensionality. Now you can apply the EquivThenEqual, use it for rewriting, etc. Thanks to Kristopher Micinski for pointing towards extensionality.

Mayer Goldberg
  • 1,378
  • 11
  • 23
  • 1
    Damn, you beat me to it ;-), I was just about to point you to the section in the FAQ – Kristopher Micinski Jun 17 '12 at 17:11
  • 2
    When your own answer is the most fitting, you can help me learn Coq by marking your answer as accepted. You'll get the reputation points. Don't be bashful--Stack Overflow is intended to work like that. – minopret Nov 11 '12 at 17:23
4

What you are looking for is called "extensionality:"

http://coq.inria.fr/V8.1/faq.html#htoc41

http://coq.inria.fr/stdlib/Coq.Logic.FunctionalExtensionality.html

http://en.wikipedia.org/wiki/Extensionality

EDIT:

You can admit predicate extensionality, as noted in the Coq FAQ.

Kristopher Micinski
  • 7,572
  • 3
  • 29
  • 34
  • I don't think this is extensionality: Notice that I'm using only one function (which serves as a context), and not two, as with extensionality. What I want is called referential transparency: If A <-> B, then I should be able to substitute B for A in any expression than any expression that contains A. On the other hand, perhaps you meant "propositional extensionality", which isn't the same, but seems to give me what I wanted. – Mayer Goldberg Jun 17 '12 at 16:51
2

This is propositional extentionality.

Lemma blah: forall (P Q: Prop), (forall (f:Prop -> Prop), f Q -> f P) -> P = Q.
  intros P Q H.
  apply (H (fun x => x = Q)).
  reflexivity.
Qed.

Section S.

Hypothesis prop_subst:
  forall (f : Prop -> Prop) (P Q : Prop), 
    (P <-> Q) -> ((f P) <-> (f Q)).

Lemma prop_subst_is_ext: forall P Q, (P <-> Q) -> P = Q.
  intros.
  apply blah.
  intro f.
  destruct (prop_subst f P Q); assumption.
Qed.

End S.

Check prop_subst_is_ext.
Haoyang Wang
  • 21
  • 1