问题
I was proving some properties of filter
and map
, everything went quite good until I stumbled on this property: filter p (map f xs) ≡ map f (filter (p ∘ f) xs)
. Here's a part of the code that's relevant:
open import Relation.Binary.PropositionalEquality
open import Data.Bool
open import Data.List hiding (filter)
import Level
filter : ∀ {a} {A : Set a} → (A → Bool) → List A → List A
filter _ [] = []
filter p (x ∷ xs) with p x
... | true = x ∷ filter p xs
... | false = filter p xs
Now, because I love writing proofs using the ≡-Reasoning
module, the first thing I tried was:
open ≡-Reasoning
open import Function
filter-map : ∀ {a b} {A : Set a} {B : Set b}
(xs : List A) (f : A → B) (p : B → Bool) →
filter p (map f xs) ≡ map f (filter (p ∘ f) xs)
filter-map [] _ _ = refl
filter-map (x ∷ xs) f p with p (f x)
... | true = begin
filter p (map f (x ∷ xs))
≡⟨ refl ⟩
f x ∷ filter p (map f xs)
-- ...
But alas, that didn't work. After trying for one hour, I finally gave up and proved it in this way:
filter-map (x ∷ xs) f p with p (f x)
... | true = cong (λ a → f x ∷ a) (filter-map xs f p)
... | false = filter-map xs f p
Still curious about why going through ≡-Reasoning
didn't work, I tried something very trivial:
filter-map-def : ∀ {a b} {A : Set a} {B : Set b}
(x : A) xs (f : A → B) (p : B → Bool) → T (p (f x)) →
filter p (map f (x ∷ xs)) ≡ f x ∷ filter p (map f xs)
filter-map-def x xs f p _ with p (f x)
filter-map-def x xs f p () | false
filter-map-def x xs f p _ | true = -- not writing refl on purpose
begin
filter p (map f (x ∷ xs))
≡⟨ refl ⟩
f x ∷ filter p (map f xs)
∎
But typechecker doesn't agree with me. It would seem that the current goal remains filter p (f x ∷ map f xs) | p (f x)
and even though I pattern matched on p (f x)
, filter
just won't reduce to f x ∷ filter p (map f xs)
.
Is there a way to make this work with ≡-Reasoning
?
Thanks!
回答1:
The trouble with with
-clauses is that Agda forgets the information it learned from pattern match unless you arrange beforehand for this information to be preserved.
More precisely, when Agda sees a with expression
clause, it replaces all the occurences of expression
in the current context and goal with a fresh variable w
and then gives you that variable with updated context and goal into the with-clause, forgetting everything about its origin.
In your case, you write filter p (map f (x ∷ xs))
inside the with-block, so it goes into scope after Agda has performed the rewriting, so Agda has already forgotten the fact that p (f x)
is true
and does not reduce the term.
You can preserve the proof of equality by using one of the "Inspect"-patterns from the standard library, but I'm not sure how it can be useful in your case.
来源:https://stackoverflow.com/questions/10320052/%e2%89%a1-reasoning-and-with-patterns