Path induction implied

Question

This is a follow-up question to Getting path induction to work in Agda

I wonder when that construct may be more expressive. It seems to me we can always express the

Answer 1

pathInd is just a dependent eliminator. Here is an isomorphic definition:

  J : ∀ {α β} {A : Set α} {x y : A}
    -> (C : {x y : A} {p : x ≡ y} -> Set β)
    -> ({x : A} -> C {x} {x})
    -> (p : x ≡ y) -> C {p = p}
  J _ b refl = b

Having this, you can define various functions on _≡_ without pattern-matching, for example:

  sym : ∀ {α} {A : Set α} {x y : A}
      -> x ≡ y
      -> y ≡ x
  sym = J (_ ≡ _) refl

  trans : ∀ {α} {A : Set α} {x y z : A}
        -> x ≡ y
        -> y ≡ z -> x ≡ z
  trans = J (_ ≡ _ -> _ ≡ _) id

  cong : ∀ {α β} {A : Set α} {B : Set β} {x y : A}
       -> (f : A -> B) 
       -> x ≡ y
       -> f x ≡ f y
  cong f = J (f _ ≡ f _) refl

  subst : ∀ {α β} {A : Set α} {x y : A}
        -> (C : A -> Set β)
        -> x ≡ y
        -> C x -> C y
  subst C = J (C _ -> C _) id

But you can't prove uniqueness of identity proofs from J as described at [1]:

  uip : ∀ {α} {A : Set α} {x y : A} -> (p q : x ≡ y) -> p ≡ q
  uip refl refl = refl

So you can express more with Agda's pattern-matching, than with just a dependent eliminator for _≡_. But you can use the --without-K option:

{-# OPTIONS --without-K #-}

open import Relation.Binary.PropositionalEquality  

uip : ∀ {α} {A : Set α} {x y : A} -> (p q : x ≡ y) -> p ≡ q
uip refl refl = refl

uip doesn't typecheck now, causing this error:

Cannot eliminate reflexive equation x = x of type A because K has
been disabled.
when checking that the pattern refl has type x ≡ x

[1] http://homotopytypetheory.org/2011/04/10/just-kidding-understanding-identity-elimination-in-homotopy-type-theory/