Proving implication (a --> c) from (b --> c) given relation between a and b

Question

I want to prove that if one implication is true (b --> c), then it follows that some other implication is also true (a --> c), given that there is an appropriate relation between the a and b.

Here is a concrete toy example that I'm trying to prove:

theory Prop imports Main
begin

lemma 
  fixes func1 :: "'a => 'a"
    and func2 :: "'a => 'a"
    and func3 :: "'a => 'a"
  assumes "∀ x y. func1 x = func1 y --> func3 x = func3 y"
      and "∀ x. func1 x = func2 x"
  shows "∀ x y. func2 x = func2 y --> func3 x = func3 y"
proof -
  from assms show ?thesis by auto
qed

end

The prove does not succed, Isabelle 2018 keeps running and the "by auto" part is colored purple. How should I go about proving this kind of lemmas?

Answer 1

It seems that your problem makes the simplifier (which is part of auto) loop. I don't really see why, but these things do happen occasionally.

When this happens, it can sometimes help to run try0 (which just tries a couple of common automatic proof methods and returns those that succeed) or sledgehammer (which tries to translate the problem into a simpler form and give it to external provers; if they can prove it, it then tries to translate the proof back to Isabelle).

In this case, both try0 and sledgehammer find that a simple apply metis can do the job. Methods like auto and simp do lots of things, including, most notably, ‘stupid’ rewriting with a pre-defined set of rules. metis is a bit more clever in what it does, but you need to manually give it every fact that it is supposed to use, and it is less adapted to Isabelle/HOL.

However, since this problem is simple first-order logic, metis can easily solve them by itself with no explicitly given facts, and it manages to avoid whatever pitfall causes auto and simp to diverge.