How does `auto` interract with biconditional (iff)

问题

I noticed, that auto is ignoring biconditionals. Here is a simplified example:

Parameter A B : Prop.
Parameter A_iff_B : A <-> B.

Theorem foo1: A -> B.
Proof.
  intros H. apply A_iff_B. assumption.
Qed.

Theorem bar1: B -> A.
Proof.
  intros H. apply A_iff_B. assumption.
Qed.

Theorem foo2_failing: A -> B.
Proof.
  intros H. auto using A_iff_B.
Abort.

Theorem bar2_failing: B -> A.
Proof.
  intros H. auto using A_iff_B.
Abort.

Now, I know that A <-> B is a syntactic sugar for A -> B /\ B -> A so I wrote two theorems to extract one or the other:

Theorem iff_forward : forall {P Q : Prop},
  (P <-> Q) -> P -> Q.
Proof.
  intros P Q H. apply H.
Qed.

Theorem iff_backward : forall {P Q : Prop},
  (P <-> Q) -> Q -> P.
Proof.
  intros P Q H. apply H.
Qed.

Theorem foo3: A -> B.
Proof.
  intros H.
  auto using (iff_forward A_iff_B).
Qed.

Theorem bar3: B -> A.
Proof.
  intros H.
  auto using (iff_backward A_iff_B).
Qed.

How come apply A_iff_B works and auto using A_iff_B does not? I thought that auto n is performing an exhaustive search of all possible sequences of apply of length <= n using the hypotheses and all theorems in a given database.
Is there a standard trick for working with biconditionals or are those two projection functions the usual solution?
Are such projection functions somewhere in the standard library? I could not found them.

回答1:

How come apply A_iff_B works and auto using A_iff_B does not?

auto generally uses simple apply instead of apply and this restricted version of apply does not handle biconditionals.

Is there a standard trick for working with biconditionals or are those two projection functions the usual solution?

You can use Hint Resolve -> (<-) feature for that:

Hint Resolve -> A_iff_B.
Hint Resolve <- A_iff_B. (* if you remove this one, then `auto` won't be able to prove the `bar3` theorem *)

Theorem foo3: A -> B.
Proof. info_auto. Qed. (* look at the output *)

Are such projection functions somewhere in the standard library?

Yes, they are called: proj1 and proj2. Here is how you can find them:

Search (?A /\ ?B -> ?A).

Or a bit easier to type, but finds a tad more stuff than we need:

Search (_ /\ _ -> _).

来源：https://stackoverflow.com/questions/47435258/how-does-auto-interract-with-biconditional-iff

标签

coq