Proof on booleans, false = true

Question

I currently am at chapter 5 of "Software foundations" but felt the need to get back to chapter one to clarify a couple of things. In particular there is an exercise I did not quite digested, in which we are asked to use destruct twice to prove a result on booleans. Here it is with names and other details changed.

Inductive bool: Type :=
|true: bool
|false: bool.

Definition fb (b1:bool) (b2:bool) : bool :=
match b1, b2 with
| false, false => false
| _, _ => true
end.

Theorem th: forall a b: bool,
  fb a b = false -> b = false.
Proof.
intros [] [] H.
- rewrite <- H. reflexivity.
- reflexivity.
- rewrite <- H. reflexivity.
- reflexivity.
Qed.

When at the first tick, context and goal are both nonsense:

H : fb true true = false
______________________________________(1/1)
true = false

Second tick the hypothesis is false. Third tick is same kind of nonsense as first one. Only fourth tick is reasonable with:

H : fb false false = false
______________________________________(1/1)
false = false

I understand that by the rewrite rules, all these things do work. However I have the impression we are quitting the narrow path of truth for the wilderness of falsity. More precisely, and AFAIK, a false hypothesis can be made to prove ANY statement, true or false. Here we use it to prove that false = true, OK why not, but still that makes me feel somewhat uncomfortable. I would not have expected a proof assistant to allow this.

Elaborating a bit

In a typical proof by contradiction, I would pick an hypothesis at random, and derive the goal till I find either a tautology or a contradiction. I would then conclude whether my hypothesis was true or false.

What happens here, in cases 1 (same for case 3), Coq starts from an hypothesis that is false:

H : fb true true = false

applies it to a goal that is a contradiction:

true = false

and combines them to find a tautology.

That is not a way of reasoning I am aware of. That recalls student 'jokes' where starting with 0=1 any absurd result on natural numbers can be proven.

Followup

So this morning during my commute I was thinking about what I had just written above. I now believe that cases 1 and 3 are proper proofs by contradiction. Indeed H is false and we use it to prove a goal that is a false. Hypotheses (values of a and b) have to be rejected. What may have confused me is that using rewrite we are doing part of the way "backward", starting from the goal.

I am a bit undecided for case 2, which reads:

H : fb true false = false
______________________________________(1/1)
false = false

which is basically false -> true, a tautology under the "principle of explosion". I would not think that could be used so directly in a proof.

Oh well, not sure I completely understood what's under the hood, but trust in Coq is untouched. Gotta go on and return to chapter 5. Thanks all for your comments.

Answer 1

First of all, thanks for providing a self-contained code.

I understand your uneasiness proving a goal using rewrite when you know that what you really ought to do is to derive a contradiction from the hypotheses. That does not make the reasoning incorrect though. It is true that under such assumptions you can prove this goal.

However I also think that this does not make the proof script really readable. In your example, you are considering all possible cases and it happens that three out of these four are impossible. When we read your proof we cannot see that. To make it clear that you are in an impossible case, there are a few tactic which are useful to say "look, I am now going to prove a contradiction to rule out this case".

One of them is exfalso. It will replace the current goal by False (since anything can be derived from False, as mentioned by @ejgallego in a comment).

A second one is absurd to say "I am now going to prove some statement and its negation" (this is basically equivalent to proving False).

A third one which is enough in your case is discriminate. It tries to find in the hypotheses a contradictory equality, such as true = false.

Theorem th: forall a b: bool,
  fb a b = false -> b = false.
Proof.
  intros [] [] H.
  - discriminate.
  - discriminate.
  - discriminate.
  - reflexivity.
Qed.

---

Now, just so you know, discriminate and reflexivity are both tried by the easy tactic. Thus the following proof will work as well (but it does not show what is going on and thus falls out of the scope of this question):

Theorem th: forall a b: bool,
  fb a b = false -> b = false.
Proof.
  intros [] [] H; easy.
Qed.

and this is syntactic sugar for the same proof:

Theorem th: forall a b: bool,
  fb a b = false -> b = false.
Proof.
  now intros [] [] H.
Qed.