How to use an unequality to simplify a if-then-else in Coq?

Question

I am in the middle of a proof, where I generated two cases by

destruct (eq_id_dec Y X)

( eq_id_dec is similar in nature to eq_nat_dec ).

This gives two cases with added assumptions e: Y = X for equality and n: Y <> X for unequality, respectively.

In the first case, I can easily use rewrite e or rewrite <- e .

But how can I efficiently make use of the unequality in the second case? Consider, eg a goal such as

(if eq_id_dec X Y then AAA else BBB) = BBB

?

I tried unfold eq_id_dec and some rewrite S but got stuck.

Answer 1

Ideally, you would like to say something like

forall (P : Prop)
       (b : {P} + {~ P})
       (H : ~ P), b = right H.

Unfortunately, it is not possible to show this result without assuming functional extensionality, because there is no useful principle that allows you to show that two proofs of negation are equal.

You can prove a generic consequence of this lemma for your case, like this:

Require Import Coq.Arith.Peano_dec.

Lemma sumboolF T P (b : {P} + {~ P}) x y : ~ P -> (if b then x else y) = y :> T.
Proof.
intros; destruct b; tauto.
Qed.

Lemma test n m : n <> m -> (if eq_nat_dec n m then 1 else 0) = 0.
Proof.
intros H; rewrite sumboolF; auto.
Qed.

This helps solving your case, but may require showing analogous results for other uses of the sumbool type.

This is one of the issues that are making us consider removing sumbool from the Software Foundations book, and just using plain booleans instead.