01 exercises
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@ -40,7 +40,10 @@ the leftmost argument only.
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```agda
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_&&'_ : Bool → Bool → Bool
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a &&' b = {!!}
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true &&' true = true
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true &&' false = false
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false &&' true = false
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false &&' false = false
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```
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One advantage of this definition is that it reads just like a Boolean truth
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@ -54,7 +57,10 @@ verbose definition.
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```agda
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_xor_ : Bool → Bool → Bool
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a xor b = {!!}
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true xor true = false
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true xor false = true
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false xor true = true
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false xor false = false
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```
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### Exercise 3 (★)
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@ -67,16 +73,18 @@ left hand side and the right hand side compute to the same value.
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```agda
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_^_ : ℕ → ℕ → ℕ
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n ^ m = {!!}
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n ^ zero = 1
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n ^ suc m = n * (n ^ m)
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^-example : 3 ^ 4 ≡ 81
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^-example = {!!} -- refl 81 should fill the hole here
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^-example = refl 81 -- refl 81 should fill the hole here
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_! : ℕ → ℕ
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n ! = {!!}
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zero ! = 1
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suc n ! = (suc n) * (n !)
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!-example : 4 ! ≡ 24
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!-example = {!!} -- refl 24 should fill the hole here
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!-example = refl 24 -- refl 24 should fill the hole here
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```
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### Exercise 4 (★)
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@ -93,10 +101,13 @@ max (suc n) (suc m) = suc (max n m)
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```agda
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min : ℕ → ℕ → ℕ
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min = {!!}
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min zero zero = zero
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min zero (suc y) = zero
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min (suc x) zero = zero
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min (suc x) (suc y) = 1 + min x y
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min-example : min 5 3 ≡ 3
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min-example = {!!} -- refl 3 should fill the hole here
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min-example = refl 3 -- refl 3 should fill the hole here
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```
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### Exercise 5 (★)
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@ -110,10 +121,11 @@ element of the list `xs` and returns the resulting list.
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```agda
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map : {X Y : Type} → (X → Y) → List X → List Y
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map f xs = {!!}
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map {X} {Y} f [] = []
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map {X} {Y} f (x :: xs) = f x :: map f xs
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map-example : map (_+ 3) (1 :: 2 :: 3 :: []) ≡ 4 :: 5 :: 6 :: []
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map-example = {!!} -- refl _ should fill the hole here
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map-example = refl _ -- refl _ should fill the hole here
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-- We write the underscore, because we don't wish to repeat
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-- the relatively long "4 :: 5 :: 6 :: []" and Agda can
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@ -130,14 +142,15 @@ should return [4 , 3 , 1], see the code below.
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```agda
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filter : {X : Type} (p : X → Bool) → List X → List X
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filter = {!!}
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filter p [] = []
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filter p (x :: xs) = if p x then x :: filter p xs else filter p xs
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is-non-zero : ℕ → Bool
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is-non-zero zero = false
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is-non-zero (suc _) = true
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filter-example : filter is-non-zero (4 :: 3 :: 0 :: 1 :: 0 :: []) ≡ 4 :: 3 :: 1 :: []
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filter-example = {!!} -- refl _ should fill the hole here
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filter-example = refl _ -- refl _ should fill the hole here
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```
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## Part II: The identity type of the Booleans (★/★★)
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@ -152,7 +165,10 @@ are the same natural number, or else is empty, if `x` and `y` are different.
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```agda
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_≣_ : Bool → Bool → Type
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a ≣ b = {!!}
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true ≣ true = 𝟙
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true ≣ false = 𝟘
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false ≣ true = 𝟘
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false ≣ false = 𝟙
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```
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### Exercise 2 (★)
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@ -161,7 +177,8 @@ a ≣ b = {!!}
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```agda
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Bool-refl : (b : Bool) → b ≣ b
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Bool-refl = {!!}
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Bool-refl true = ⋆
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Bool-refl false = ⋆
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```
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### Exercise 3 (★★)
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@ -175,10 +192,16 @@ back and forth between `a ≣ b` and `a ≡ b`.
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```agda
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≡-to-≣ : (a b : Bool) → a ≡ b → a ≣ b
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≡-to-≣ = {!!}
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≡-to-≣ true true x = ⋆
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≡-to-≣ true false ()
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≡-to-≣ false true ()
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≡-to-≣ false false x = ⋆
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≣-to-≡ : (a b : Bool) → a ≣ b → a ≡ b
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≣-to-≡ = {!!}
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≣-to-≡ true true ⋆ = refl _
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≣-to-≡ true false ()
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≣-to-≡ false true ()
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≣-to-≡ false false ⋆ = refl _
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```
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## Part III: Proving in Agda (★★/★★★)
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@ -199,7 +222,10 @@ Use pattern matching to **prove** that `||` is commutative.
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```agda
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||-is-commutative : (a b : Bool) → a || b ≡ b || a
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||-is-commutative a b = {!!}
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||-is-commutative true true = refl _
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||-is-commutative true false = refl _
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||-is-commutative false true = refl _
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||-is-commutative false false = refl _
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```
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### Exercise 2 (★★)
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@ -208,8 +234,11 @@ Taking inspiration from the above, try to **state** and **prove** that `&&` is
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commutative.
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```agda
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&&-is-commutative : {!!}
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&&-is-commutative = {!!}
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&&-is-commutative : (a b : Bool) → a && b ≡ b && a
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&&-is-commutative true true = refl _
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&&-is-commutative true false = refl _
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&&-is-commutative false true = refl _
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&&-is-commutative false false = refl _
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```
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### Exercise 3 (★★)
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@ -218,10 +247,24 @@ commutative.
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```agda
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&&-is-associative : (a b c : Bool) → a && (b && c) ≡ (a && b) && c
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&&-is-associative = {!!}
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&&-is-associative true true true = refl _
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&&-is-associative true true false = refl _
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&&-is-associative true false true = refl _
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&&-is-associative true false false = refl _
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&&-is-associative false true true = refl _
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&&-is-associative false true false = refl _
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&&-is-associative false false true = refl _
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&&-is-associative false false false = refl _
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&&'-is-associative : (a b c : Bool) → a &&' (b &&' c) ≡ (a &&' b) &&' c
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&&'-is-associative = {!!}
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&&'-is-associative true true true = refl _
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&&'-is-associative true true false = refl _
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&&'-is-associative true false true = refl _
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&&'-is-associative true false false = refl _
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&&'-is-associative false true true = refl _
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&&'-is-associative false true false = refl _
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&&'-is-associative false false true = refl _
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&&'-is-associative false false false = refl _
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```
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**Question**: Can you spot a downside of the more verbose definition of `&&'`
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@ -249,7 +292,10 @@ max-is-commutative (suc n) (suc m) = to-show
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```agda
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min-is-commutative : (n m : ℕ) → min n m ≡ min m n
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min-is-commutative = {!!}
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min-is-commutative zero zero = refl _
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min-is-commutative zero (suc m) = refl _
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min-is-commutative (suc n) zero = refl _
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min-is-commutative (suc n) (suc m) = ap suc (min-is-commutative n m)
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```
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### Exercise 5 (★★★)
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@ -259,7 +305,8 @@ number `n`.
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```agda
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0-right-neutral : (n : ℕ) → n ≡ n + 0
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0-right-neutral = {!!}
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0-right-neutral zero = refl _
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0-right-neutral (suc n) = ap suc (0-right-neutral n)
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```
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### Exercise 6 (★★★)
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@ -271,9 +318,11 @@ Try to **prove** these equations using pattern matching and inductive proofs.
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```agda
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map-id : {X : Type} (xs : List X) → map id xs ≡ xs
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map-id xs = {!!}
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map-id {X} [] = refl _
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map-id {X} (x :: xs) = ap (λ xs → x :: xs) (map-id xs)
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map-comp : {X Y Z : Type} (f : X → Y) (g : Y → Z)
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(xs : List X) → map (g ∘ f) xs ≡ map g (map f xs)
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map-comp f g xs = {!!}
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map-comp {X} {Y} {Z} f g [] = refl _
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map-comp {X} {Y} {Z} f g (x :: xs) = ap (λ xs → g (f x) :: xs) (map-comp f g xs)
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```
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