Lists: Lists and higher-order functions

module plfa.part1.Lists where

This chapter discusses the list data type. It gives further examples of many of the techniques we have developed so far, and provides examples of polymorphic types and higher-order functions.

Imports

import Relation.Binary.PropositionalEquality as Eq
open Eq using (_≡_; refl; sym; trans; cong)
open Eq.≡-Reasoning
open import Data.Bool using (Bool; true; false; T; _∧_; _∨_; not)
open import Data.Nat using (ℕ; zero; suc; _+_; _*_; _∸_; _≤_; s≤s; z≤n)
open import Data.Nat.Properties using
  (+-assoc; +-identityˡ; +-identityʳ; *-assoc; *-identityˡ; *-identityʳ)
open import Relation.Nullary using (¬_; Dec; yes; no)
open import Data.Product using (_×_; ∃; ∃-syntax) renaming (_,_ to ⟨_,_⟩)
open import Function using (_∘_)
open import Level using (Level)
open import plfa.part1.Isomorphism using (_≃_; _⇔_)

Lists

Lists are defined in Agda as follows:

data List (A : Set) : Set where
  []  : List A
  _∷_ : A → List A → List A

infixr 5 _∷_

Let’s unpack this definition. If A is a set, then List A is a set. The next two lines tell us that [] (pronounced nil) is a list of type A (often called the empty list), and that _∷_ (pronounced cons, short for constructor) takes a value of type A and a value of type List A and returns a value of type List A. Operator _∷_ has precedence level 5 and associates to the right.

For example,

_ : List ℕ
_ = 0 ∷ 1 ∷ 2 ∷ []

denotes the list of the first three natural numbers. Since _∷_ associates to the right, the term parses as 0 ∷ (1 ∷ (2 ∷ [])). Here 0 is the first element of the list, called the head, and 1 ∷ (2 ∷ []) is a list of the remaining elements, called the tail. A list is a strange beast: it has a head and a tail, nothing in between, and the tail is itself another list!

As we’ve seen, parameterised types can be translated to indexed types. The definition above is equivalent to the following:

data List′ : Set → Set where
  []′  : ∀ {A : Set} → List′ A
  _∷′_ : ∀ {A : Set} → A → List′ A → List′ A

Each constructor takes the parameter as an implicit argument. Thus, our example list could also be written:

_ : List ℕ
_ = _∷_ {ℕ} 0 (_∷_ {ℕ} 1 (_∷_ {ℕ} 2 ([] {ℕ})))

where here we have provided the implicit parameters explicitly.

Including the pragma:

{-# BUILTIN LIST List #-}

tells Agda that the type List corresponds to the Haskell type list, and the constructors [] and _∷_ correspond to nil and cons respectively, allowing a more efficient representation of lists.

List syntax

We can write lists more conveniently by introducing the following definitions:

pattern [_] z = z ∷ []
pattern [_,_] y z = y ∷ z ∷ []
pattern [_,_,_] x y z = x ∷ y ∷ z ∷ []
pattern [_,_,_,_] w x y z = w ∷ x ∷ y ∷ z ∷ []
pattern [_,_,_,_,_] v w x y z = v ∷ w ∷ x ∷ y ∷ z ∷ []
pattern [_,_,_,_,_,_] u v w x y z = u ∷ v ∷ w ∷ x ∷ y ∷ z ∷ []

This is our first use of pattern declarations. For instance, the third line tells us that [ x , y , z ] is equivalent to x ∷ y ∷ z ∷ [], and permits the former to appear either in a pattern on the left-hand side of an equation, or a term on the right-hand side of an equation.

Append

Our first function on lists is written _++_ and pronounced append:

infixr 5 _++_

_++_ : ∀ {A : Set} → List A → List A → List A
[]       ++ ys  =  ys
(x ∷ xs) ++ ys  =  x ∷ (xs ++ ys)

The type A is an implicit argument to append, making it a polymorphic function (one that can be used at many types). The empty list appended to another list yields the other list. A non-empty list appended to another list yields a list with head the same as the head of the first list and tail the same as the tail of the first list appended to the second list.

Here is an example, showing how to compute the result of appending two lists:

_ : [ 0 , 1 , 2 ] ++ [ 3 , 4 ] ≡ [ 0 , 1 , 2 , 3 , 4 ]
_ =
  begin
    0 ∷ 1 ∷ 2 ∷ [] ++ 3 ∷ 4 ∷ []
  ≡⟨⟩
    0 ∷ (1 ∷ 2 ∷ [] ++ 3 ∷ 4 ∷ [])
  ≡⟨⟩
    0 ∷ 1 ∷ (2 ∷ [] ++ 3 ∷ 4 ∷ [])
  ≡⟨⟩
    0 ∷ 1 ∷ 2 ∷ ([] ++ 3 ∷ 4 ∷ [])
  ≡⟨⟩
    0 ∷ 1 ∷ 2 ∷ 3 ∷ 4 ∷ []
  ∎

Appending two lists requires time linear in the number of elements in the first list.

Reasoning about append

We can reason about lists in much the same way that we reason about numbers. Here is the proof that append is associative:

++-assoc : ∀ {A : Set} (xs ys zs : List A)
  → (xs ++ ys) ++ zs ≡ xs ++ (ys ++ zs)
++-assoc [] ys zs =
  begin
    ([] ++ ys) ++ zs
  ≡⟨⟩
    ys ++ zs
  ≡⟨⟩
    [] ++ (ys ++ zs)
  ∎
++-assoc (x ∷ xs) ys zs =
  begin
    (x ∷ xs ++ ys) ++ zs
  ≡⟨⟩
    x ∷ (xs ++ ys) ++ zs
  ≡⟨⟩
    x ∷ ((xs ++ ys) ++ zs)
  ≡⟨ cong (x ∷_) (++-assoc xs ys zs) ⟩
    x ∷ (xs ++ (ys ++ zs))
  ≡⟨⟩
    x ∷ xs ++ (ys ++ zs)
  ∎

The proof is by induction on the first argument. The base case instantiates to [], and follows by straightforward computation. The inductive case instantiates to x ∷ xs, and follows by straightforward computation combined with the inductive hypothesis. As usual, the inductive hypothesis is indicated by a recursive invocation of the proof, in this case ++-assoc xs ys zs.

Recall that Agda supports sections. Applying cong (x ∷_) promotes the inductive hypothesis:

(xs ++ ys) ++ zs ≡ xs ++ (ys ++ zs)

to the equality:

x ∷ ((xs ++ ys) ++ zs) ≡ x ∷ (xs ++ (ys ++ zs))

which is needed in the proof.

It is also easy to show that [] is a left and right identity for _++_. That it is a left identity is immediate from the definition:

++-identityˡ : ∀ {A : Set} (xs : List A) → [] ++ xs ≡ xs
++-identityˡ xs =
  begin
    [] ++ xs
  ≡⟨⟩
    xs
  ∎

That it is a right identity follows by simple induction:

++-identityʳ : ∀ {A : Set} (xs : List A) → xs ++ [] ≡ xs
++-identityʳ [] =
  begin
    [] ++ []
  ≡⟨⟩
    []
  ∎
++-identityʳ (x ∷ xs) =
  begin
    (x ∷ xs) ++ []
  ≡⟨⟩
    x ∷ (xs ++ [])
  ≡⟨ cong (x ∷_) (++-identityʳ xs) ⟩
    x ∷ xs
  ∎

As we will see later, these three properties establish that _++_ and [] form a monoid over lists.

Length

Our next function finds the length of a list:

length : ∀ {A : Set} → List A → ℕ
length []        =  zero
length (x ∷ xs)  =  suc (length xs)

Again, it takes an implicit parameter A. The length of the empty list is zero. The length of a non-empty list is one greater than the length of the tail of the list.

Here is an example showing how to compute the length of a list:

_ : length [ 0 , 1 , 2 ] ≡ 3
_ =
  begin
    length (0 ∷ 1 ∷ 2 ∷ [])
  ≡⟨⟩
    suc (length (1 ∷ 2 ∷ []))
  ≡⟨⟩
    suc (suc (length (2 ∷ [])))
  ≡⟨⟩
    suc (suc (suc (length {ℕ} [])))
  ≡⟨⟩
    suc (suc (suc zero))
  ∎

Computing the length of a list requires time linear in the number of elements in the list.

In the second-to-last line, we cannot write simply length [] but must instead write length {ℕ} []. Since [] has no elements, Agda has insufficient information to infer the implicit parameter.

Reasoning about length

The length of one list appended to another is the sum of the lengths of the lists:

length-++ : ∀ {A : Set} (xs ys : List A)
  → length (xs ++ ys) ≡ length xs + length ys
length-++ {A} [] ys =
  begin
    length ([] ++ ys)
  ≡⟨⟩
    length ys
  ≡⟨⟩
    length {A} [] + length ys
  ∎
length-++ (x ∷ xs) ys =
  begin
    length ((x ∷ xs) ++ ys)
  ≡⟨⟩
    suc (length (xs ++ ys))
  ≡⟨ cong suc (length-++ xs ys) ⟩
    suc (length xs + length ys)
  ≡⟨⟩
    length (x ∷ xs) + length ys
  ∎

The proof is by induction on the first argument. The base case instantiates to [], and follows by straightforward computation. As before, Agda cannot infer the implicit type parameter to length, and it must be given explicitly. The inductive case instantiates to x ∷ xs, and follows by straightforward computation combined with the inductive hypothesis. As usual, the inductive hypothesis is indicated by a recursive invocation of the proof, in this case length-++ xs ys, and it is promoted by the congruence cong suc.

Reverse

Using append, it is easy to formulate a function to reverse a list:

reverse : ∀ {A : Set} → List A → List A
reverse []        =  []
reverse (x ∷ xs)  =  reverse xs ++ [ x ]

The reverse of the empty list is the empty list. The reverse of a non-empty list is the reverse of its tail appended to a unit list containing its head.

Here is an example showing how to reverse a list:

_ : reverse [ 0 , 1 , 2 ] ≡ [ 2 , 1 , 0 ]
_ =
  begin
    reverse (0 ∷ 1 ∷ 2 ∷ [])
  ≡⟨⟩
    reverse (1 ∷ 2 ∷ []) ++ [ 0 ]
  ≡⟨⟩
    (reverse (2 ∷ []) ++ [ 1 ]) ++ [ 0 ]
  ≡⟨⟩
    ((reverse [] ++ [ 2 ]) ++ [ 1 ]) ++ [ 0 ]
  ≡⟨⟩
    (([] ++ [ 2 ]) ++ [ 1 ]) ++ [ 0 ]
  ≡⟨⟩
    (([] ++ 2 ∷ []) ++ 1 ∷ []) ++ 0 ∷ []
  ≡⟨⟩
    (2 ∷ [] ++ 1 ∷ []) ++ 0 ∷ []
  ≡⟨⟩
    2 ∷ ([] ++ 1 ∷ []) ++ 0 ∷ []
  ≡⟨⟩
    (2 ∷ 1 ∷ []) ++ 0 ∷ []
  ≡⟨⟩
    2 ∷ (1 ∷ [] ++ 0 ∷ [])
  ≡⟨⟩
    2 ∷ 1 ∷ ([] ++ 0 ∷ [])
  ≡⟨⟩
    2 ∷ 1 ∷ 0 ∷ []
  ≡⟨⟩
    [ 2 , 1 , 0 ]
  ∎

Reversing a list in this way takes time quadratic in the length of the list. This is because reverse ends up appending lists of lengths 1, 2, up to n - 1, where n is the length of the list being reversed, append takes time linear in the length of the first list, and the sum of the numbers up to n - 1 is n * (n - 1) / 2. (We will validate that last fact in an exercise later in this chapter.)

Exercise `reverse-++-commute` (recommended)

Show that the reverse of one list appended to another is the reverse of the second appended to the reverse of the first:

postulate
  reverse-++-commute : ∀ {A : Set} {xs ys : List A}
    → reverse (xs ++ ys) ≡ reverse ys ++ reverse xs

Exercise `reverse-involutive` (recommended)

A function is an involution if when applied twice it acts as the identity function. Show that reverse is an involution:

postulate
  reverse-involutive : ∀ {A : Set} {xs : List A}
    → reverse (reverse xs) ≡ xs

Faster reverse

The definition above, while easy to reason about, is less efficient than one might expect since it takes time quadratic in the length of the list. The idea is that we generalise reverse to take an additional argument:

shunt : ∀ {A : Set} → List A → List A → List A
shunt []       ys  =  ys
shunt (x ∷ xs) ys  =  shunt xs (x ∷ ys)

The definition is by recursion on the first argument. The second argument actually becomes larger, but this is not a problem because the argument on which we recurse becomes smaller.

Shunt is related to reverse as follows:

shunt-reverse : ∀ {A : Set} (xs ys : List A)
  → shunt xs ys ≡ reverse xs ++ ys
shunt-reverse [] ys =
  begin
    shunt [] ys
  ≡⟨⟩
    ys
  ≡⟨⟩
    reverse [] ++ ys
  ∎
shunt-reverse (x ∷ xs) ys =
  begin
    shunt (x ∷ xs) ys
  ≡⟨⟩
    shunt xs (x ∷ ys)
  ≡⟨ shunt-reverse xs (x ∷ ys) ⟩
    reverse xs ++ (x ∷ ys)
  ≡⟨⟩
    reverse xs ++ ([ x ] ++ ys)
  ≡⟨ sym (++-assoc (reverse xs) [ x ] ys) ⟩
    (reverse xs ++ [ x ]) ++ ys
  ≡⟨⟩
    reverse (x ∷ xs) ++ ys
  ∎

The proof is by induction on the first argument. The base case instantiates to [], and follows by straightforward computation. The inductive case instantiates to x ∷ xs and follows by the inductive hypothesis and associativity of append. When we invoke the inductive hypothesis, the second argument actually becomes larger, but this is not a problem because the argument on which we induct becomes smaller.

Generalising on an auxiliary argument, which becomes larger as the argument on which we recurse or induct becomes smaller, is a common trick. It belongs in your quiver of arrows, ready to slay the right problem.

Having defined shunt be generalisation, it is now easy to respecialise to give a more efficient definition of reverse:

reverse′ : ∀ {A : Set} → List A → List A
reverse′ xs = shunt xs []

Given our previous lemma, it is straightforward to show the two definitions equivalent:

reverses : ∀ {A : Set} (xs : List A)
  → reverse′ xs ≡ reverse xs
reverses xs =
  begin
    reverse′ xs
  ≡⟨⟩
    shunt xs []
  ≡⟨ shunt-reverse xs [] ⟩
    reverse xs ++ []
  ≡⟨ ++-identityʳ (reverse xs) ⟩
    reverse xs
  ∎

Here is an example showing fast reverse of the list [ 0 , 1 , 2 ]:

_ : reverse′ [ 0 , 1 , 2 ] ≡ [ 2 , 1 , 0 ]
_ =
  begin
    reverse′ (0 ∷ 1 ∷ 2 ∷ [])
  ≡⟨⟩
    shunt (0 ∷ 1 ∷ 2 ∷ []) []
  ≡⟨⟩
    shunt (1 ∷ 2 ∷ []) (0 ∷ [])
  ≡⟨⟩
    shunt (2 ∷ []) (1 ∷ 0 ∷ [])
  ≡⟨⟩
    shunt [] (2 ∷ 1 ∷ 0 ∷ [])
  ≡⟨⟩
    2 ∷ 1 ∷ 0 ∷ []
  ∎

Now the time to reverse a list is linear in the length of the list.

Map

Map applies a function to every element of a list to generate a corresponding list. Map is an example of a higher-order function, one which takes a function as an argument or returns a function as a result:

map : ∀ {A B : Set} → (A → B) → List A → List B
map f []        =  []
map f (x ∷ xs)  =  f x ∷ map f xs

Map of the empty list is the empty list. Map of a non-empty list yields a list with head the same as the function applied to the head of the given list, and tail the same as map of the function applied to the tail of the given list.

Here is an example showing how to use map to increment every element of a list:

_ : map suc [ 0 , 1 , 2 ] ≡ [ 1 , 2 , 3 ]
_ =
  begin
    map suc (0 ∷ 1 ∷ 2 ∷ [])
  ≡⟨⟩
    suc 0 ∷ map suc (1 ∷ 2 ∷ [])
  ≡⟨⟩
    suc 0 ∷ suc 1 ∷ map suc (2 ∷ [])
  ≡⟨⟩
    suc 0 ∷ suc 1 ∷ suc 2 ∷ map suc []
  ≡⟨⟩
    suc 0 ∷ suc 1 ∷ suc 2 ∷ []
  ≡⟨⟩
    1 ∷ 2 ∷ 3 ∷ []
  ∎

Map requires time linear in the length of the list.

It is often convenient to exploit currying by applying map to a function to yield a new function, and at a later point applying the resulting function:

sucs : List ℕ → List ℕ
sucs = map suc

_ : sucs [ 0 , 1 , 2 ] ≡ [ 1 , 2 , 3 ]
_ =
  begin
    sucs [ 0 , 1 , 2 ]
  ≡⟨⟩
    map suc [ 0 , 1 , 2 ]
  ≡⟨⟩
    [ 1 , 2 , 3 ]
  ∎

Any type that is parameterised on another type, such as lists, has a corresponding map, which accepts a function and returns a function from the type parameterised on the domain of the function to the type parameterised on the range of the function. Further, a type that is parameterised on n types will have a map that is parameterised on n functions.

Exercise `map-compose` (practice)

Prove that the map of a composition is equal to the composition of two maps:

postulate
  map-compose : ∀ {A B C : Set} {f : A → B} {g : B → C}
    → map (g ∘ f) ≡ map g ∘ map f

The last step of the proof requires extensionality.

Exercise `map-++-commute` (practice)

Prove the following relationship between map and append:

postulate
  map-++-commute : ∀ {A B : Set} {f : A → B} {xs ys : List A}
   →  map f (xs ++ ys) ≡ map f xs ++ map f ys

Exercise `map-Tree` (practice)

Define a type of trees with leaves of type A and internal nodes of type B:

data Tree (A B : Set) : Set where
  leaf : A → Tree A B
  node : Tree A B → B → Tree A B → Tree A B

Define a suitable map operator over trees:

postulate
  map-Tree : ∀ {A B C D : Set}
    → (A → C) → (B → D) → Tree A B → Tree C D

Fold

Fold takes an operator and a value, and uses the operator to combine each of the elements of the list, taking the given value as the result for the empty list:

foldr : ∀ {A B : Set} → (A → B → B) → B → List A → B
foldr _⊗_ e []        =  e
foldr _⊗_ e (x ∷ xs)  =  x ⊗ foldr _⊗_ e xs

Fold of the empty list is the given value. Fold of a non-empty list uses the operator to combine the head of the list and the fold of the tail of the list.

Here is an example showing how to use fold to find the sum of a list:

_ : foldr _+_ 0 [ 1 , 2 , 3 , 4 ] ≡ 10
_ =
  begin
    foldr _+_ 0 (1 ∷ 2 ∷ 3 ∷ 4 ∷ [])
  ≡⟨⟩
    1 + foldr _+_ 0 (2 ∷ 3 ∷ 4 ∷ [])
  ≡⟨⟩
    1 + (2 + foldr _+_ 0 (3 ∷ 4 ∷ []))
  ≡⟨⟩
    1 + (2 + (3 + foldr _+_ 0 (4 ∷ [])))
  ≡⟨⟩
    1 + (2 + (3 + (4 + foldr _+_ 0 [])))
  ≡⟨⟩
    1 + (2 + (3 + (4 + 0)))
  ∎

Fold requires time linear in the length of the list.

It is often convenient to exploit currying by applying fold to an operator and a value to yield a new function, and at a later point applying the resulting function:

sum : List ℕ → ℕ
sum = foldr _+_ 0

_ : sum [ 1 , 2 , 3 , 4 ] ≡ 10
_ =
  begin
    sum [ 1 , 2 , 3 , 4 ]
  ≡⟨⟩
    foldr _+_ 0 [ 1 , 2 , 3 , 4 ]
  ≡⟨⟩
    10
  ∎

Just as the list type has two constructors, [] and _∷_, so the fold function takes two arguments, e and _⊗_ (in addition to the list argument). In general, a data type with n constructors will have a corresponding fold function that takes n arguments.

Exercise `product` (recommended)

Use fold to define a function to find the product of a list of numbers. For example:

product [ 1 , 2 , 3 , 4 ] ≡ 24

-- Your code goes here

Exercise `foldr-++` (recommended)

Show that fold and append are related as follows:

postulate
  foldr-++ : ∀ {A B : Set} (_⊗_ : A → B → B) (e : B) (xs ys : List A) →
    foldr _⊗_ e (xs ++ ys) ≡ foldr _⊗_ (foldr _⊗_ e ys) xs

Exercise `map-is-foldr` (practice)

Show that map can be defined using fold:

postulate
  map-is-foldr : ∀ {A B : Set} {f : A → B} →
    map f ≡ foldr (λ x xs → f x ∷ xs) []

This requires extensionality.

Exercise `fold-Tree` (practice)

Define a suitable fold function for the type of trees given earlier:

postulate
  fold-Tree : ∀ {A B C : Set}
    → (A → C) → (C → B → C → C) → Tree A B → C

-- Your code goes here

Exercise `map-is-fold-Tree` (practice)

Demonstrate an analogue of map-is-foldr for the type of trees.

-- Your code goes here

Exercise `sum-downFrom` (stretch)

Define a function that counts down as follows:

downFrom : ℕ → List ℕ
downFrom zero     =  []
downFrom (suc n)  =  n ∷ downFrom n

For example:

_ : downFrom 3 ≡ [ 2 , 1 , 0 ]
_ = refl

Prove that the sum of the numbers (n - 1) + ⋯ + 0 is equal to n * (n ∸ 1) / 2:

postulate
  sum-downFrom : ∀ (n : ℕ)
    → sum (downFrom n) * 2 ≡ n * (n ∸ 1)

Monoids

Typically when we use a fold the operator is associative and the value is a left and right identity for the operator, meaning that the operator and the value form a monoid.

We can define a monoid as a suitable record type:

record IsMonoid {A : Set} (_⊗_ : A → A → A) (e : A) : Set where
  field
    assoc : ∀ (x y z : A) → (x ⊗ y) ⊗ z ≡ x ⊗ (y ⊗ z)
    identityˡ : ∀ (x : A) → e ⊗ x ≡ x
    identityʳ : ∀ (x : A) → x ⊗ e ≡ x

open IsMonoid

As examples, sum and zero, multiplication and one, and append and the empty list, are all examples of monoids:

+-monoid : IsMonoid _+_ 0
+-monoid =
  record
    { assoc = +-assoc
    ; identityˡ = +-identityˡ
    ; identityʳ = +-identityʳ
    }

*-monoid : IsMonoid _*_ 1
*-monoid =
  record
    { assoc = *-assoc
    ; identityˡ = *-identityˡ
    ; identityʳ = *-identityʳ
    }

++-monoid : ∀ {A : Set} → IsMonoid {List A} _++_ []
++-monoid =
  record
    { assoc = ++-assoc
    ; identityˡ = ++-identityˡ
    ; identityʳ = ++-identityʳ
    }

If _⊗_ and e form a monoid, then we can re-express fold on the same operator and an arbitrary value:

foldr-monoid : ∀ {A : Set} (_⊗_ : A → A → A) (e : A) → IsMonoid _⊗_ e →
  ∀ (xs : List A) (y : A) → foldr _⊗_ y xs ≡ foldr _⊗_ e xs ⊗ y
foldr-monoid _⊗_ e ⊗-monoid [] y =
  begin
    foldr _⊗_ y []
  ≡⟨⟩
    y
  ≡⟨ sym (identityˡ ⊗-monoid y) ⟩
    (e ⊗ y)
  ≡⟨⟩
    foldr _⊗_ e [] ⊗ y
  ∎
foldr-monoid _⊗_ e ⊗-monoid (x ∷ xs) y =
  begin
    foldr _⊗_ y (x ∷ xs)
  ≡⟨⟩
    x ⊗ (foldr _⊗_ y xs)
  ≡⟨ cong (x ⊗_) (foldr-monoid _⊗_ e ⊗-monoid xs y) ⟩
    x ⊗ (foldr _⊗_ e xs ⊗ y)
  ≡⟨ sym (assoc ⊗-monoid x (foldr _⊗_ e xs) y) ⟩
    (x ⊗ foldr _⊗_ e xs) ⊗ y
  ≡⟨⟩
    foldr _⊗_ e (x ∷ xs) ⊗ y
  ∎

As a consequence, using a previous exercise, we have the following:

foldr-monoid-++ : ∀ {A : Set} (_⊗_ : A → A → A) (e : A) → IsMonoid _⊗_ e →
  ∀ (xs ys : List A) → foldr _⊗_ e (xs ++ ys) ≡ foldr _⊗_ e xs ⊗ foldr _⊗_ e ys
foldr-monoid-++ _⊗_ e monoid-⊗ xs ys =
  begin
    foldr _⊗_ e (xs ++ ys)
  ≡⟨ foldr-++ _⊗_ e xs ys ⟩
    foldr _⊗_ (foldr _⊗_ e ys) xs
  ≡⟨ foldr-monoid _⊗_ e monoid-⊗ xs (foldr _⊗_ e ys) ⟩
    foldr _⊗_ e xs ⊗ foldr _⊗_ e ys
  ∎

Exercise `foldl` (practice)

Define a function foldl which is analogous to foldr, but where operations associate to the left rather than the right. For example:

foldr _⊗_ e [ x , y , z ]  =  x ⊗ (y ⊗ (z ⊗ e))
foldl _⊗_ e [ x , y , z ]  =  ((e ⊗ x) ⊗ y) ⊗ z

-- Your code goes here

Exercise `foldr-monoid-foldl` (practice)

Show that if _⊗_ and e form a monoid, then foldr _⊗_ e and foldl _⊗_ e always compute the same result.

-- Your code goes here

All

We can also define predicates over lists. Two of the most important are All and Any.

Predicate All P holds if predicate P is satisfied by every element of a list:

data All {A : Set} (P : A → Set) : List A → Set where
  []  : All P []
  _∷_ : ∀ {x : A} {xs : List A} → P x → All P xs → All P (x ∷ xs)

The type has two constructors, reusing the names of the same constructors for lists. The first asserts that P holds for every element of the empty list. The second asserts that if P holds of the head of a list and for every element of the tail of a list, then P holds for every element of the list. Agda uses types to disambiguate whether the constructor is building a list or evidence that All P holds.

For example, All (_≤ 2) holds of a list where every element is less than or equal to two. Recall that z≤n proves zero ≤ n for any n, and that if m≤n proves m ≤ n then s≤s m≤n proves suc m ≤ suc n, for any m and n:

_ : All (_≤ 2) [ 0 , 1 , 2 ]
_ = z≤n ∷ s≤s z≤n ∷ s≤s (s≤s z≤n) ∷ []

Here _∷_ and [] are the constructors of All P rather than of List A. The three items are proofs of 0 ≤ 2, 1 ≤ 2, and 2 ≤ 2, respectively.

(One might wonder whether a pattern such as [_,_,_] can be used to construct values of type All as well as type List, since both use the same constructors. Indeed it can, so long as both types are in scope when the pattern is declared. That’s not the case here, since List is defined before [_,_,_], but All is defined later.)

Any

Predicate Any P holds if predicate P is satisfied by some element of a list:

data Any {A : Set} (P : A → Set) : List A → Set where
  here  : ∀ {x : A} {xs : List A} → P x → Any P (x ∷ xs)
  there : ∀ {x : A} {xs : List A} → Any P xs → Any P (x ∷ xs)

The first constructor provides evidence that the head of the list satisfies P, while the second provides evidence that some element of the tail of the list satisfies P. For example, we can define list membership as follows:

infix 4 _∈_ _∉_

_∈_ : ∀ {A : Set} (x : A) (xs : List A) → Set
x ∈ xs = Any (x ≡_) xs

_∉_ : ∀ {A : Set} (x : A) (xs : List A) → Set
x ∉ xs = ¬ (x ∈ xs)

For example, zero is an element of the list [ 0 , 1 , 0 , 2 ]. Indeed, we can demonstrate this fact in two different ways, corresponding to the two different occurrences of zero in the list, as the first element and as the third element:

_ : 0 ∈ [ 0 , 1 , 0 , 2 ]
_ = here refl

_ : 0 ∈ [ 0 , 1 , 0 , 2 ]
_ = there (there (here refl))

Further, we can demonstrate that three is not in the list, because any possible proof that it is in the list leads to contradiction:

not-in : 3 ∉ [ 0 , 1 , 0 , 2 ]
not-in (here ())
not-in (there (here ()))
not-in (there (there (here ())))
not-in (there (there (there (here ()))))
not-in (there (there (there (there ()))))

The five occurrences of () attest to the fact that there is no possible evidence for 3 ≡ 0, 3 ≡ 1, 3 ≡ 0, 3 ≡ 2, and 3 ∈ [], respectively.

All and append

A predicate holds for every element of one list appended to another if and only if it holds for every element of both lists:

All-++-⇔ : ∀ {A : Set} {P : A → Set} (xs ys : List A) →
  All P (xs ++ ys) ⇔ (All P xs × All P ys)
All-++-⇔ xs ys =
  record
    { to       =  to xs ys
    ; from     =  from xs ys
    }
  where

  to : ∀ {A : Set} {P : A → Set} (xs ys : List A) →
    All P (xs ++ ys) → (All P xs × All P ys)
  to [] ys Pys = ⟨ [] , Pys ⟩
  to (x ∷ xs) ys (Px ∷ Pxs++ys) with to xs ys Pxs++ys
  ... | ⟨ Pxs , Pys ⟩ = ⟨ Px ∷ Pxs , Pys ⟩

  from : ∀ { A : Set} {P : A → Set} (xs ys : List A) →
    All P xs × All P ys → All P (xs ++ ys)
  from [] ys ⟨ [] , Pys ⟩ = Pys
  from (x ∷ xs) ys ⟨ Px ∷ Pxs , Pys ⟩ =  Px ∷ from xs ys ⟨ Pxs , Pys ⟩

Exercise `Any-++-⇔` (recommended)

Prove a result similar to All-++-⇔, but with Any in place of All, and a suitable replacement for _×_. As a consequence, demonstrate an equivalence relating _∈_ and _++_.

-- Your code goes here

Exercise `All-++-≃` (stretch)

Show that the equivalence All-++-⇔ can be extended to an isomorphism.

-- Your code goes here

Exercise `¬Any≃All¬` (stretch)

First generalise composition to arbitrary levels, using universe polymorphism:

_∘′_ : ∀ {ℓ₁ ℓ₂ ℓ₃ : Level} {A : Set ℓ₁} {B : Set ℓ₂} {C : Set ℓ₃}
  → (B → C) → (A → B) → A → C
(g ∘′ f) x  =  g (f x)

Show that Any and All satisfy a version of De Morgan’s Law:

postulate
  ¬Any≃All¬ : ∀ {A : Set} (P : A → Set) (xs : List A)
    → (¬_ ∘′ Any P) xs ≃ All (¬_ ∘′ P) xs

Do we also have the following?

postulate
  ¬All≃Any¬ : ∀ {A : Set} (P : A → Set) (xs : List A)
    → (¬_ ∘′ All P) xs ≃ Any (¬_ ∘′ P) xs

If so, prove; if not, explain why.

Decidability of All

If we consider a predicate as a function that yields a boolean, it is easy to define an analogue of All, which returns true if a given predicate returns true for every element of a list:

all : ∀ {A : Set} → (A → Bool) → List A → Bool
all p  =  foldr _∧_ true ∘ map p

The function can be written in a particularly compact style by using the higher-order functions map and foldr.

As one would hope, if we replace booleans by decidables there is again an analogue of All. First, return to the notion of a predicate P as a function of type A → Set, taking a value x of type A into evidence P x that a property holds for x. Say that a predicate P is decidable if we have a function that for a given x can decide P x:

Decidable : ∀ {A : Set} → (A → Set) → Set
Decidable {A} P  =  ∀ (x : A) → Dec (P x)

Then if predicate P is decidable, it is also decidable whether every element of a list satisfies the predicate:

All? : ∀ {A : Set} {P : A → Set} → Decidable P → Decidable (All P)
All? P? []                                 =  yes []
All? P? (x ∷ xs) with P? x   | All? P? xs
...                 | yes Px | yes Pxs     =  yes (Px ∷ Pxs)
...                 | no ¬Px | _           =  no λ{ (Px ∷ Pxs) → ¬Px Px   }
...                 | _      | no ¬Pxs     =  no λ{ (Px ∷ Pxs) → ¬Pxs Pxs }

If the list is empty, then trivially P holds for every element of the list. Otherwise, the structure of the proof is similar to that showing that the conjunction of two decidable propositions is itself decidable, using _∷_ rather than ⟨_,_⟩ to combine the evidence for the head and tail of the list.

Exercise `any?` (stretch)

Just as All has analogues all and All? which determine whether a predicate holds for every element of a list, so does Any have analogues any and Any? which determine whether a predicate holds for some element of a list. Give their definitions.

-- Your code goes here

Exercise `All-∀` (practice)

Show that All P xs is isomorphic to ∀ {x} → x ∈ xs → P x.

-- You code goes here

Exercise `Any-∃` (practice)

Show that Any P xs is isomorphic to ∃[ x ∈ xs ] P x.

-- You code goes here

Exercise `filter?` (stretch)

Define the following variant of the traditional filter function on lists, which given a decidable predicate and a list returns all elements of the list satisfying the predicate:

postulate
  filter? : ∀ {A : Set} {P : A → Set}
    → (P? : Decidable P) → List A → ∃[ ys ]( All P ys )

Standard Library

Definitions similar to those in this chapter can be found in the standard library:

import Data.List using (List; _++_; length; reverse; map; foldr; downFrom)
import Data.List.All using (All; []; _∷_)
import Data.List.Any using (Any; here; there)
import Data.List.Membership.Propositional using (_∈_)
import Data.List.Properties
  using (reverse-++-commute; map-compose; map-++-commute; foldr-++)
  renaming (mapIsFold to map-is-foldr)
import Algebra.Structures using (IsMonoid)
import Relation.Unary using (Decidable)
import Relation.Binary using (Decidable)

The standard library version of IsMonoid differs from the one given here, in that it is also parameterised on an equivalence relation.

Both Relation.Unary and Relation.Binary define a version of Decidable, one for unary relations (as used in this chapter where P ranges over unary predicates) and one for binary relations (as used earlier, where _≤_ ranges over a binary relation).

Unicode

This chapter uses the following unicode:

∷  U+2237  PROPORTION  (\::)
⊗  U+2297  CIRCLED TIMES  (\otimes, \ox)
∈  U+2208  ELEMENT OF  (\in)
∉  U+2209  NOT AN ELEMENT OF  (\inn, \notin)

Imports

Lists

List syntax

Append

Reasoning about append

Length

Reasoning about length

Reverse

Exercise reverse-++-commute (recommended)

Exercise reverse-involutive (recommended)

Faster reverse

Map

Exercise map-compose (practice)

Exercise map-++-commute (practice)

Exercise map-Tree (practice)

Fold

Exercise product (recommended)

Exercise foldr-++ (recommended)

Exercise map-is-foldr (practice)

Exercise fold-Tree (practice)

Exercise map-is-fold-Tree (practice)

Exercise sum-downFrom (stretch)