我怎么说服coq（A / \\ B）/ \\ C == A / \\ B / \\ C？

Question

In my proof I stumble upon problems where there is an A /\\ B /\\ C in my assumptions, and I need to prove (A /\\ B) /\\ C . These are logically exactly the same, but coq won't solve these with assumption. .

I have been solving these by applying an axiom, but is there a more elegant (and correct) way to handle this?

Answer 1

So the way I've gone about it is by defining my lemma,

Lemma conj_assoc : forall A B C, A /\ (B /\ C) <-> (A /\ B) /\ C.

That is one implies the other.

intros. split. will then split this into two goals.

1. A /\\ (B /\\ C) -> (A /\\ B) /\\ C
2. (A /\\ B) /\\ C -> A /\\ (B /\\ C)

Proving each of these is roughly the same. For (1),

• intro Habc. to get the assumption from the left hand size.
• destruct Habc as [Ha Hbc]. destruct Hbc as [Hb Hc]. to get the individual assumptions.
• auto to use these assumptions.

I leave it to you to work out (2) but it is very similar.

Then Qed.

Answer 2

If you have A /\\ B /\\ C as an assumption, and your goal is (A /\\ B) /\\ C , you can use the tactic tauto . This tactic solves all tautologies in the propositional calculus. There is also a tactic firstorder which can solve some formulas with quantifiers.

If you have A /\\ B /\\ C and you'd like to pass (A /\\ B) /\\ C as an argument to a lemma, you'll need to work a bit more. One method is to set (A /\\ B) /\\ C as an intermediate goal and prove it:

assert ((A /\ B) /\ C). tauto.

If A , B and C are large expressions, you can use a compound tactic to match over the hypothesis H : A /\\ B /\\ C and apply the tauto tactic to it. This is a heavy-handed approach, overkill in this case, but useful in more complex situations where you want to automate a proof with many similar cases.

match type of H with ?x /\ ?y /\ ?z =>
  assert (x /\ (y /\ z)); [tauto | clear H]
end.

There's an easier way, which is to apply a known lemma that performs the transformation.

apply and_assoc in H.

You can find the lemma by browsing the library documentation. You can also search for it. This isn't the easiest lemma to search for because it's an equivalence and the search tools are geared towards implications and equalities. You can use SearchPattern (_ /\\ _ /\\ _). to look for lemmas of the form forall x1 … xn, ?A /\\ ?B /\\ ?C (where ?A , ?B and ?C can be any expression). You can use SearchRewrite (_ /\\ _ /\\ _) to look for lemmas of the form forall x1 … xn, (?A /\\ ?B /\\ ?C) = ?D . Unfortunately, this doesn't find what we're after, which is a lemma of the form forall x1 … xn, (?A /\\ ?B /\\ ?C) <-> ?D . What does work is

Coq < SearchPattern (_ <-> (_ /\ _ /\ _))
and_assoc: forall A B C : Prop, (A /\ B) /\ C <-> A /\ B /\ C

Answer 3

As a general tip, if you have something like this that you suspect to be obvious, check the standard library. Here's how: Locate "/\\". produces a response that resolves the Notation for us,

Notation            Scope     
"A /\ B" := and A B  : type_scope
                      (default interpretation)

Now we can issue the command, SearchAbout and. to see what is in scope, and find that and_assoc witnesses the implication you are interested in. In fact, you can take a cue from your intuition: the intuition tactic can take advantage of this implication on its own.

Lemma conj_example : forall A B C D,
  (A /\ B) /\ C -> (A /\ (B /\ C) -> D) -> D.
Proof. intuition. Qed.