I'm writing Agda code as I read the HoTT book. I'm stuck on Lemma 2.3.9:
data _≡_ {X : Set} : X -> X -> Set where
refl : {x : X} -> x ≡ x
infix 4 _≡_
-- Lemma 2.1.2
_·_ : {A : Set} {x y z : A} -> x ≡ y -> y ≡ z -> x ≡ z
refl · refl = refl
-- Lemma 2.3.1
transp : {A : Set} {P : A -> Set} {x y : A} -> x ≡ y -> P x -> P y
transp refl f = f
lemma2'3'9 : {A : Set}{P : A -> Set}{x y z : A}{p : x ≡ y}{q : y ≡ z}{u : P x} ->
(transp q (transp p u)) ≡ (transp (p · q) u)
lemma2'3'9 {p = refl} {q = refl} = ?
Type-checking with Adga Emacs Mode gives me the following error:
?0 : transp refl (transp refl u) ≡ transp (refl · refl) u
_X_53 : Set [ at /home/user/prog/agda/sample.agda:12,38-39 ]
———— Errors ————————————————————————————————————————————————
Failed to solve the following constraints:
P x =< _X_53 (blocked on _X_53)
Questions
- What is '_X_53', and why is it greater than or equal to (P x)?
- How can I get rid of this error?
Note I wrote a working example of Lemma 2.3.9 in Coq, so I'm assuming it's possible in Agda.
Inductive eq {X:Type} (x: X) : X -> Type :=
| refl : eq x x.
Notation "x = y" := (eq x y)
(at level 70, no associativity)
: type_scope.
Definition eqInd{A} (C: forall x y: A, x = y -> Type) (c: forall x: A, C x x (refl x)) (x y: A): forall p: x = y, C x y p :=
fun xy: x = y => match xy with
| refl _ => c x
end.
Definition dot'{A}{x y: A}: x = y -> forall z: A, y = z -> x = z :=
let D := fun x y: A => fun p: x = y => forall z: A, forall q: y = z, x = z in
let d: forall x, D x x (refl x) := let E: forall x z: A, forall q: x = z, Type := fun x z: A => fun q: x = z => x = z in
let e := fun x => refl x
in fun x z => fun q => eqInd E e x z q
in fun p: x = y => eqInd D d x y p.
(* Lemma 2.1.2 *)
Definition dot{A}{x y z: A}: x = y -> y = z -> x = z :=
fun p: x = y => dot' p z.
Definition id {A} := fun a: A => a.
(* Lemma 2.3.1 *)
Definition transp{A} {P: A -> Type} {x y: A}: x = y -> P x -> P y :=
fun p =>
let D := fun x y: A => fun p: x = y => P x -> P y in
let d: forall x, D x x (refl x) := fun x => id
in eqInd D d x y p.
Lemma L_2_3_9{A}{P: A -> Type}{x y z: A}{p: x = y}{q: y = z}{u: P x}:
transp q (transp p u) = transp (dot p q) u.
Proof.
unfold transp, dot, dot'.
rewrite <- q.
rewrite <- p.
reflexivity.
Qed.