Understanding specialize tactic

Question

Trying to comprehend the answer of @keep_learning I walked through this code step by step:

Inductive nostutter {X:Type} : list X -> Prop :=
| ns_nil : nostutter []
| ns_one : forall (x : X), nostutter [x]
| ns_cons: forall (x : X) (h : X) (t : list X), nostutter (h::t) -> x <> h -> nostutter (x::h::t).

Example test_nostutter_4: not (nostutter [3;1;1;4]).
Proof.
  intro.
  inversion_clear H.
  inversion_clear H0.
  unfold not in H2.
  (* We are here *)
  specialize (H2 eq_refl).
  apply H2.
Qed.

Here is what we have before excuting specialize

H1 : 3 <> 1
H : nostutter [1; 4]
H2 : 1 = 1 -> False
============================
False

Here is eq Prop whose constructor eq_refl is used in specialize:

Inductive eq (A:Type) (x:A) : A -> Prop :=
    eq_refl : x = x :>A

where "x = y :> A" := (@eq A x y) : type_scope.

I can't explain, how this command works:

specialize (H2 eq_refl).

I read about specialize in reference manual, but the explanation there is too broad. As far as I understand it sees that "1 = 1" expression in H2 satisfies eq_refl constructor and therefore eq proposition is True. Then it simplifies the expression:

True -> False => False

And we get

H1 : 3 <> 1
H : nostutter [1; 4]
H2 : False
============================
False

Can somebody provide me a minimal example with explanation of what is specialize doing, so I could freely use it?

Update

Trying to imitate how specialize works using apply I did the following:

Example specialize {A B: Type} (H: A -> B) (a: A): B.
Proof.
  apply H in a.

This gives:

A : Type
B : Type
H : A -> B
a : B
============================
B

Almost the same as specialize, only different hypothesis name.

In test_nostutter_4 theorem I tried this and it worked:

remember (@eq_refl nat 1) as Heq.
apply H2 in Heq as H3.

It gives us:

H1 : 3 <> 1
H : nostutter [1; 4]
H2 : 1 = 1 -> False
Heq : 1 = 1
H3 : False
HeqHeq : Heq = eq_refl
============================
False

This one was more complex, we had to introduce a new hypothesis Heq. But we got what we need - H3 at the end.

Does specialize internally use something like remember? Or is it possible to solve it with apply but without remember?

Answer 1

specialize, in its simplest form, simply replaces a given hypothesis with that hypothesis applied to some other term.

In this proof,

Example specialize {A B: Type} (H: A -> B) (a: A): B.
Proof.
  specialize (H a).
  exact H.
Qed.

we initially have the hypothesis H: A -> B. When we call specialize (H a), we apply H to a (apply as in function application). This gives us something of type B. specialize then gets rid of the old H for us and replaces it with the result of the application. It gives the new hypothesis the same name: H.

In your case, we have H2: 1 = 1 -> False, which is a function from the type 1 = 1 to the type False. That means that H2 applied to eq_refl is of type False, i.e. H2 eq_refl: False. When we use the tactic specialize (H2 eq_refl)., the old H2 is cleared and replaced by a new term (H2 eq_refl) whose type is False. It keeps the old name H2, though.

specialize is useful when you're sure that you're only going to use a hypothesis once, since it automatically gets rid of the old hypothesis. One disadvantage is that the old name may not fit the meaning of the new hypothesis. However, in your case and in my example, H is a generic enough name that it works either way.

---

To your update...

specialize is a core tactic defined directly in the ltac plugin. It doesn't use any other tactic internally, since it is its internals.

If you want to keep a hypothesis, you can use the as modifier, which works for both specialize and apply. In the proof

Example specialize {A B: Type} (H: A -> B) (a: A): B.
Proof.

if you do specialize (H a) as H0., instead of clearing H, it'll introduce a new hypothesis H0: B. apply H in a as H0. has the same effect.