Applying a Program Definition fails with "unable to unify Prop with [goal]" - coq

Question

In Coq, I showed the associativity of append on vectors using:
Require Import Coq.Vectors.VectorDef Omega.
Program Definition t_app_assoc v p q r (a : t v p) (b : t v q) (c : t v r) :=
append (append a b) c = append a (append b c).
Next Obligation. omega. Qed.
I now want to apply this equality in a proof. Below is the easiest goal that I would expect to be provable with t_app_assoc. Of course it can be proven by simpl - this is just an example.
Goal (append (append (nil nat) (nil _)) (nil _)
= append (nil _) (append (nil _) (nil _))).
apply t_app_assoc.
This apply fails with:
Error: Unable to unify "Prop" with
"append (append (nil nat) (nil nat)) (nil nat) =
append (nil nat) (append (nil nat) (nil nat))".
How can I apply t_app_assoc? Or is there a better way to define it? I thought I needed a Program Definition, because simply using a Lemma leads to a type error because t v (p + (q + r)) and t v (p + q + r) are not the same to Coq.

Answer 1

Prologue
I guess what you want to to is to prove that the vector concatenation is associative and then use that fact as a lemma.
But t_app_assoc as you define it has the following type:
t_app_assoc
: forall (v : Type) (p q r : nat), t v p -> t v q -> t v r -> Prop
You basically want to use : instead of := as follows.
From Coq Require Import Vector Arith.
Import VectorNotations.
Import EqNotations. (* rew notation, see below *)
Section Append.
Variable A : Type.
Variable p q r : nat.
Variables (a : t A p) (b : t A q) (c : t A r).
Fail Lemma t_app_assoc :
append (append a b) c = append a (append b c).
Unfortunately, we cannot even state a lemma like this using the usual homogeneous equality.
The left-hand side has the following type:
Check append (append a b) c : t A (p + q + r).
whereas the right-hand side is of type
Check append a (append b c) : t A (p + (q + r)).
Since t A (p + q + r) is not the same as t A (p + (q + r)) we cannot use = to state the above lemma.
Let me describe some ways of working around this issue:
rew notation
Lemma t_app_assoc_rew :
append (append a b) c = rew (plus_assoc _ _ _) in
append a (append b c).
Admitted.
Here we just use the law of associativity of addition for natural numbers to cast the type of RHS to t A (p + q + r).
To make it work one needs to Import EqNotations. before.
cast function
This is a common problem, so the authors of the Vector library decided to provide a cast function with the following type:
cast :
forall (A : Type) (m : nat),
Vector.t A m -> forall n : nat, m = n -> Vector.t A n
Let me show how one can use it to prove the law of associativity for vectors. But let's prove the following auxiliary lemma first:
Lemma uncast {X n} {v : Vector.t X n} e :
cast v e = v.
Proof. induction v as [|??? IH]; simpl; rewrite ?IH; reflexivity. Qed.
Now we are all set:
Lemma t_app_assoc_cast (a : t A p) (b : t A q) (c : t A r) :
append (append a b) c = cast (append a (append b c)) (plus_assoc _ _ _).
Proof.
generalize (Nat.add_assoc p q r).
induction a as [|h p' a' IH]; intros e.
- now rewrite uncast.
- simpl; f_equal. apply IH.
Qed.
Heterogeneous equality (a.k.a. John Major equality)
Lemma t_app_assoc_jmeq :
append (append a b) c ~= append a (append b c).
Admitted.
End Append.
If you compare the definition of the homogeneous equality
Inductive eq (A : Type) (x : A) : A -> Prop :=
eq_refl : x = x.
and the definition of heterogeneous equality
Inductive JMeq (A : Type) (x : A) : forall B : Type, B -> Prop :=
JMeq_refl : x ~= x.
you will see that with JMeq the LHS and RHS don't have to be of the same type and this is why the statement of t_app_assoc_jmeq looks a bit simpler than the previous ones.
Other approaches to vectors
See e.g. this question
and this one;
I also find this answer
very useful too.