Explanation transitivity of equal Coq

Question

I have a proof which is already proved.

Lemma Equal_Trans : forall T : Type, forall y x z : T,
      Equal x y -> Equal y z -> Equal x z.
forall (T : Type) (y x z : T),
Equal x y -> Equal y z -> Equal x z

And secondly, I have the correction of the addition commutativity proof.

Lemma Add_com : forall x x' : Nat, Equal (Add x x') (Add x' x).
forall x x' : Nat, Equal (Add x x') (Add x' x)
Proof.
forall x x' : Nat, Equal (Add x x') (Add x' x)
  intros.
x, x': Nat
Equal (Add x x') (Add x' x) induction x.
x': Nat
Equal (Add 0 x') (Add x' 0)
x: nat
x': Nat
IHx: Equal (Add x x') (Add x' x)
Equal (Add (S x) x') (Add x' (S x))
  -
x': Nat
Equal (Add 0 x') (Add x' 0) simpl.
x': Nat
Equal x' (Add x' 0) apply Add_zero.
  -
x: nat
x': Nat
IHx: Equal (Add x x') (Add x' x)
Equal (Add (S x) x') (Add x' (S x)) simpl.
x: nat
x': Nat
IHx: Equal (Add x x') (Add x' x)
Equal (S (Add x x')) (Add x' (S x))
    apply (Equal_Trans Nat (S (Add x' x))). (* var y *)
x: nat
x': Nat
IHx: Equal (Add x x') (Add x' x)
Equal (S (Add x x')) (S (Add x' x))
x: nat
x': Nat
IHx: Equal (Add x x') (Add x' x)
Equal (S (Add x' x)) (Add x' (S x))
    +
x: nat
x': Nat
IHx: Equal (Add x x') (Add x' x)
Equal (S (Add x x')) (S (Add x' x)) apply Equal_Morph.
x: nat
x': Nat
IHx: Equal (Add x x') (Add x' x)
Equal (Add x x') (Add x' x) assumption.
    +
x: nat
x': Nat
IHx: Equal (Add x x') (Add x' x)
Equal (S (Add x' x)) (Add x' (S x)) apply Add_S.
Qed.

However, I don't understand the use of Equal_Trans (line 6). If I understand, Equal_Trans takes 3 arguments: y x z? But why there is only 1 argument using Equal_Trans in Add_com lemma?

Thank you in advance for your help.

Answer

The tactic apply tries to fill in the blanks to match the type of the provided term with the goal. In this case, the goal (when the tactic is used) is probably something along the lines of a = b (based on your followup, it's actually Equal a b). The type of Equal_Trans (when all the arguments are used) is x = z (Equal x z), so to unify these two types, we should have x := a and z := b. That only leaves y as ambiguous, so we have to provide it.

To address your followup, no, Equal_Trans does not take just one argument. It takes a type (Nat in your case) three elements of that type (y, x and z) and two equality proofs. However, remember that functions in Coq are curried, which means that you can call them with fewer arguments, but the result will be a function of the remaining arguments.

So really, when we say apply (Equal_Trans Nat (S (Add x' x))., we're saying "take this thing that has type forall (x z: Nat), Equal x (S (Add x' x)) -> Equal (S (Add x' x)) z -> Equal x z and try to fill in some of the arguments to match it with my goal".

Coq looks at that type and realizes that the goal already looks like Equal x z, so it's able to deduce what x and z have to be. Equal_Trans still takes two more arguments that Coq can't figure out on its own (the proofs of Equal x y and Equal y z), so that's what the rest of the proof is doing.

We use transitivity with y := S (Add x' x) because we can prove Equal (S (Add x x')) (S (Add x' x)) using the inductive hypothesis (IHx). We can also prove Equal (S (Add x' x)) (Add x' (S x)) by using the definition of Add. Hence, it's natural to route the proof of equality through S (Add x' x).

Now, we don't have to use transitivity with y := S (Add x' x). We could prove that (S (Add x x')) and (Add x' (S x)) are both equal to some other element of Nat. But the easiest and most direct route is through S (Add x' x).