[Agda] How can I prove associativity of vectors?

[Conor McBride]

Hi

A little bird told me I should pay attention for a change. Here’s what I cooked up.

Given that intensional equality is always a disappointment, it is too much to expect
the values on either side of a propositional equation to have intensionally equal
types. The method for working with heterogeneous equality is always to work with
whole telescopes and make sure that whenever the types stretch apart (like the
vector types below), you always have an equation which tells you why they are
provably equal.

I could probably be more systematic about building the “resp” family, but it’s late and
I’m tired.

All the best

Conor

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module VVV where

data Nat : Set where
 ze : Nat
 su : Nat -> Nat

_+_ : Nat -> Nat -> Nat
ze + y = y
su x + y = su (x + y)

data Vec (X : Set) : Nat -> Set where
 [] : Vec X ze
 _,_ : {n : Nat} -> X -> Vec X n -> Vec X (su n)

data _==_ {A : Set}(a : A) : {B : Set} -> B -> Set where
 refl : a == a

_++_ : {X : Set}{m n : Nat} -> Vec X m -> Vec X n -> Vec X (m + n)
[] ++ ys = ys
(x , xs) ++ ys = x , (xs ++ ys)

record _*_ (S T : Set) : Set where
 constructor _,_
 field
   fst : S
   snd : T
open _*_ public

resp1 : {A : Set}{B : A -> Set}
 (f : (a : A) -> B a)
 {a a' : A} -> a == a' ->
 f a == f a'
resp1 f refl = refl

resp3 : {A : Set}{B : A -> Set}{C : (a : A)(b : B a) -> Set}{D : (a : A)(b : B a)(c : C a b) -> Set}
 (f : (a : A)(b : B a)(c : C a b) -> D a b c) ->
 {a a' : A} -> a == a' ->
 {b : B a}{b' : B a'} -> b == b' ->
 {c : C a b}{c' : C a' b'} -> c == c' ->
 f a b c == f a' b' c'
resp3 f refl refl refl = refl

vc : {X : Set}(n : Nat)(x : X) -> Vec X n -> Vec X (su n)
vc n x xs = x , xs

vass : {X : Set}{l m n : Nat}(xs : Vec X l)(ys : Vec X m)(zs : Vec X n) ->
 (((l + m) + n) == (l + (m + n))) *
 (((xs ++ ys) ++ zs) == (xs ++ (ys ++ zs)))
vass [] ys zs = refl , refl
vass (x , xs) ys zs with vass xs ys zs
... | q , q' = (resp1 su q) , resp3 vc q refl q'
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