问题
As the title asks, I wish for an example where:
Section Question:
Definition A: Prop := <whatever you like>.
Definition B:Prop := <whatever you like>.
Definition/Inductive/Fixpoint P: Prop -> Type := <whatever you like>.
Theorem AEquivB: A <-> B.
Proof. <supply proof here>. Qed.
(* Question 1. can we pick a P, A, B to prove this? *)
Theorem PA_not_equals_Pb: P A <> P B.
Proof. <supply proof here>. Qed.
(* Question 1.5. can we pick a P, A, B to prove this? *)
Theorem PA_not_equiv_PB: ~(P A <-> P B)
Proof. <supply proof here>. Qed.
In general, I am interested to understand whether "proof equivalence" is "good enough" to be used as "equality" in a sense, or whether there are situations where we can have P A, and A <-> B, but not P B.
回答1:
It is consistent with Coq that forall A B : Prop, (A <-> B) -> A = B. (That is, you can add this as an axiom and the theory won't collapse.) This axiom is called propositional extensionality. As A = B quickly gives forall P : Prop -> Prop, P A <-> P B, there are no terms P, A, B such that (A <-> B) /\ ~(P A <-> P B), since this would contradict the axiom, but we know it is consistent. Similarly, we also quickly get P A = P B, which means we cannot also get P A <> P B. Note that even though such P, A, B that violate propositional extensionality do not exist, we still cannot prove propositional extensionality. Coq simply doesn't have the strength to talk about itself like that (which is good, since that means you can customize it), which is why propositional extensionality needs to be added as an axiom if you want it.
来源：https://stackoverflow.com/questions/61242486/provide-example-in-coq-where-a-b-prop-p-prop-type-such-that-a-b-bu