Decomposing equality of constructors coq

Question

Often in Coq I find myself doing the following: I have the proof goal, for example:

some_constructor a c d = some_constructor b c d

And I r

Answer 1

You can use Coq's searching capabilities:

  Search (?X _ = ?X _).
  Search (_ _ = _ _).

Among some noise it reveals a lemma

f_equal: forall (A B : Type) (f : A -> B) (x y : A), x = y -> f x = f y

And its siblings for multi-argument equalities: f_equal2 ... f_equal5 (as of Coq version 8.4).

Here is an example:

Inductive silly : Set :=
  | some_constructor : nat -> nat -> nat -> silly
  | another_constructor : nat -> nat -> silly.

Goal forall x y,
    x = 42 ->
    y = 6 * 7 ->
    some_constructor x 0 1 = some_constructor y 0 1.
  intros x y Hx Hy.
  apply f_equal3; try reflexivity.

At this point all you need to prove is x = y.

Answer 2

In particular, standard Coq provides the f_equal tactic.

Inductive u : Type := U : nat -> nat -> nat -> u.

Lemma U1 x y z1 z2 : U x y z1 = U x y z2.
f_equal

Also, ssreflect provides a general-purpose congruence tactic congr.