# Quantifiers: Universals and existentials

module plfa.part1.Quantifiers where

This chapter introduces universal and existential quantification.

## Imports

import Relation.Binary.PropositionalEquality as Eq open Eq using (_≡_; refl) open import Data.Nat using (ℕ; zero; suc; _+_; _*_) open import Relation.Nullary using (¬_) open import Data.Product using (_×_; proj₁; proj₂) renaming (_,_ to ⟨_,_⟩) open import Data.Sum using (_⊎_; inj₁; inj₂) open import plfa.part1.Isomorphism using (_≃_; extensionality; ∀-extensionality) open import Function using (_∘_)

## Universals

We formalise universal quantification using the dependent function type, which has appeared throughout this book. For instance, in Chapter Induction we showed addition is associative:

`+-assoc : ∀ (m n p : ℕ) → (m + n) + p ≡ m + (n + p)`

which asserts for all natural numbers `m`

, `n`

, and `p`

that `(m + n) + p ≡ m + (n + p)`

holds. It is a dependent function, which given values for `m`

, `n`

, and `p`

returns evidence for the corresponding equation.

In general, given a variable `x`

of type `A`

and a proposition `B x`

which contains `x`

as a free variable, the universally quantified proposition `∀ (x : A) → B x`

holds if for every term `M`

of type `A`

the proposition `B M`

holds. Here `B M`

stands for the proposition `B x`

with each free occurrence of `x`

replaced by `M`

. Variable `x`

appears free in `B x`

but bound in `∀ (x : A) → B x`

.

Evidence that `∀ (x : A) → B x`

holds is of the form

`λ (x : A) → N x`

where `N x`

is a term of type `B x`

, and `N x`

and `B x`

both contain a free variable `x`

of type `A`

. Given a term `L`

providing evidence that `∀ (x : A) → B x`

holds, and a term `M`

of type `A`

, the term `L M`

provides evidence that `B M`

holds. In other words, evidence that `∀ (x : A) → B x`

holds is a function that converts a term `M`

of type `A`

into evidence that `B M`

holds.

`∀ (x : A) → B x`

holds and that `M`

is a term of type `A`

then we may conclude that `B M`

holds:∀-elim : ∀ {A : Set} {B : A → Set} → (L : ∀ (x : A) → B x) → (M : A) ----------------- → B M ∀-elim L M = L M

As with `→-elim`

, the rule corresponds to function application.

Functions arise as a special case of dependent functions, where the range does not depend on a variable drawn from the domain. When a function is viewed as evidence of implication, both its argument and result are viewed as evidence, whereas when a dependent function is viewed as evidence of a universal, its argument is viewed as an element of a data type and its result is viewed as evidence of a proposition that depends on the argument. This difference is largely a matter of interpretation, since in Agda a value of a type and evidence of a proposition are indistinguishable.

Dependent function types are sometimes referred to as dependent products, because if `A`

is a finite type with values `x₁ , ⋯ , xₙ`

, and if each of the types `B x₁ , ⋯ , B xₙ`

has `m₁ , ⋯ , mₙ`

distinct members, then `∀ (x : A) → B x`

has `m₁ * ⋯ * mₙ`

members. Indeed, sometimes the notation `∀ (x : A) → B x`

is replaced by a notation such as `Π[ x ∈ A ] (B x)`

, where `Π`

stands for product. However, we will stick with the name dependent function, because (as we will see) dependent product is ambiguous.

#### Exercise `∀-distrib-×`

(recommended)

Show that universals distribute over conjunction:postulate ∀-distrib-× : ∀ {A : Set} {B C : A → Set} → (∀ (x : A) → B x × C x) ≃ (∀ (x : A) → B x) × (∀ (x : A) → C x)

Compare this with the result (`→-distrib-×`

) in Chapter Connectives.

Hint: you will need to use `∀-extensionality`

.

#### Exercise `⊎∀-implies-∀⊎`

(practice)

Show that a disjunction of universals implies a universal of disjunctions:postulate ⊎∀-implies-∀⊎ : ∀ {A : Set} {B C : A → Set} → (∀ (x : A) → B x) ⊎ (∀ (x : A) → C x) → ∀ (x : A) → B x ⊎ C x

Does the converse hold? If so, prove; if not, explain why.

#### Exercise `∀-×`

(practice)

Consider the following type.data Tri : Set where aa : Tri bb : Tri cc : Tri

Let `B`

be a type indexed by `Tri`

, that is `B : Tri → Set`

. Show that `∀ (x : Tri) → B x`

is isomorphic to `B aa × B bb × B cc`

.

Hint: you will need to use `∀-extensionality`

.

## Existentials

Given a variable `x`

of type `A`

and a proposition `B x`

which contains `x`

as a free variable, the existentially quantified proposition `Σ[ x ∈ A ] B x`

holds if for some term `M`

of type `A`

the proposition `B M`

holds. Here `B M`

stands for the proposition `B x`

with each free occurrence of `x`

replaced by `M`

. Variable `x`

appears free in `B x`

but bound in `Σ[ x ∈ A ] B x`

.

data Σ (A : Set) (B : A → Set) : Set where ⟨_,_⟩ : (x : A) → B x → Σ A BWe define a convenient syntax for existentials as follows:

Σ-syntax = Σ infix 2 Σ-syntax syntax Σ-syntax A (λ x → Bx) = Σ[ x ∈ A ] Bx

This is our first use of a syntax declaration, which specifies that the term on the left may be written with the syntax on the right. The special syntax is available only when the identifier `Σ-syntax`

is imported.

Evidence that `Σ[ x ∈ A ] B x`

holds is of the form `⟨ M , N ⟩`

where `M`

is a term of type `A`

, and `N`

is evidence that `B M`

holds.

record Σ′ (A : Set) (B : A → Set) : Set where field proj₁′ : A proj₂′ : B proj₁′

Here record construction

```
record
{ proj₁′ = M
; proj₂′ = N
}
```

corresponds to the term

`⟨ M , N ⟩`

where `M`

is a term of type `A`

and `N`

is a term of type `B M`

.

Products arise as a special case of existentials, where the second component does not depend on a variable drawn from the first component. When a product is viewed as evidence of a conjunction, both of its components are viewed as evidence, whereas when it is viewed as evidence of an existential, the first component is viewed as an element of a datatype and the second component is viewed as evidence of a proposition that depends on the first component. This difference is largely a matter of interpretation, since in Agda a value of a type and evidence of a proposition are indistinguishable.

Existentials are sometimes referred to as dependent sums, because if `A`

is a finite type with values `x₁ , ⋯ , xₙ`

, and if each of the types `B x₁ , ⋯ B xₙ`

has `m₁ , ⋯ , mₙ`

distinct members, then `Σ[ x ∈ A ] B x`

has `m₁ + ⋯ + mₙ`

members, which explains the choice of notation for existentials, since `Σ`

stands for sum.

Existentials are sometimes referred to as dependent products, since products arise as a special case. However, that choice of names is doubly confusing, since universals also have a claim to the name dependent product and since existentials also have a claim to the name dependent sum.

A common notation for existentials is`∃`

(analogous to `∀`

for universals). We follow the convention of the Agda standard library, and reserve this notation for the case where the domain of the bound variable is left implicit:∃ : ∀ {A : Set} (B : A → Set) → Set ∃ {A} B = Σ A B ∃-syntax = ∃ syntax ∃-syntax (λ x → B) = ∃[ x ] B

The special syntax is available only when the identifier `∃-syntax`

is imported. We will tend to use this syntax, since it is shorter and more familiar.

`∀ x → B x → C`

holds, where `C`

does not contain `x`

as a free variable, and given evidence that `∃[ x ] B x`

holds, we may conclude that `C`

holds:∃-elim : ∀ {A : Set} {B : A → Set} {C : Set} → (∀ x → B x → C) → ∃[ x ] B x --------------- → C ∃-elim f ⟨ x , y ⟩ = f x y

In other words, if we know for every `x`

of type `A`

that `B x`

implies `C`

, and we know for some `x`

of type `A`

that `B x`

holds, then we may conclude that `C`

holds. This is because we may instantiate that proof that `∀ x → B x → C`

to any value `x`

of type `A`

and any `y`

of type `B x`

, and exactly such values are provided by the evidence for `∃[ x ] B x`

.

∀∃-currying : ∀ {A : Set} {B : A → Set} {C : Set} → (∀ x → B x → C) ≃ (∃[ x ] B x → C) ∀∃-currying = record { to = λ{ f → λ{ ⟨ x , y ⟩ → f x y }} ; from = λ{ g → λ{ x → λ{ y → g ⟨ x , y ⟩ }}} ; from∘to = λ{ f → refl } ; to∘from = λ{ g → extensionality λ{ ⟨ x , y ⟩ → refl }} }

The result can be viewed as a generalisation of currying. Indeed, the code to establish the isomorphism is identical to what we wrote when discussing implication.

#### Exercise `∃-distrib-⊎`

(recommended)

Show that existentials distribute over disjunction:postulate ∃-distrib-⊎ : ∀ {A : Set} {B C : A → Set} → ∃[ x ] (B x ⊎ C x) ≃ (∃[ x ] B x) ⊎ (∃[ x ] C x)

#### Exercise `∃×-implies-×∃`

(practice)

Show that an existential of conjunctions implies a conjunction of existentials:postulate ∃×-implies-×∃ : ∀ {A : Set} {B C : A → Set} → ∃[ x ] (B x × C x) → (∃[ x ] B x) × (∃[ x ] C x)

Does the converse hold? If so, prove; if not, explain why.

#### Exercise `∃-⊎`

(practice)

Let `Tri`

and `B`

be as in Exercise `∀-×`

. Show that `∃[ x ] B x`

is isomorphic to `B aa ⊎ B bb ⊎ B cc`

.

## An existential example

Recall the definitions of`even`

and `odd`

from Chapter Relations:data even : ℕ → Set data odd : ℕ → Set data even where even-zero : even zero even-suc : ∀ {n : ℕ} → odd n ------------ → even (suc n) data odd where odd-suc : ∀ {n : ℕ} → even n ----------- → odd (suc n)

A number is even if it is zero or the successor of an odd number, and odd if it is the successor of an even number.

We will show that a number is even if and only if it is twice some other number, and odd if and only if it is one more than twice some other number. In other words, we will show:

`even n`

iff `∃[ m ] ( m * 2 ≡ n)`

`odd n`

iff `∃[ m ] (1 + m * 2 ≡ n)`

By convention, one tends to write constant factors first and to put the constant term in a sum last. Here we’ve reversed each of those conventions, because doing so eases the proof.

Here is the proof in the forward direction:even-∃ : ∀ {n : ℕ} → even n → ∃[ m ] ( m * 2 ≡ n) odd-∃ : ∀ {n : ℕ} → odd n → ∃[ m ] (1 + m * 2 ≡ n) even-∃ even-zero = ⟨ zero , refl ⟩ even-∃ (even-suc o) with odd-∃ o ... | ⟨ m , refl ⟩ = ⟨ suc m , refl ⟩ odd-∃ (odd-suc e) with even-∃ e ... | ⟨ m , refl ⟩ = ⟨ m , refl ⟩

We define two mutually recursive functions. Given evidence that `n`

is even or odd, we return a number `m`

and evidence that `m * 2 ≡ n`

or `1 + m * 2 ≡ n`

. We induct over the evidence that `n`

is even or odd:

If the number is even because it is zero, then we return a pair consisting of zero and the evidence that twice zero is zero.

If the number is even because it is one more than an odd number, then we apply the induction hypothesis to give a number

`m`

and evidence that`1 + m * 2 ≡ n`

. We return a pair consisting of`suc m`

and evidence that`suc m * 2 ≡ suc n`

, which is immediate after substituting for`n`

.If the number is odd because it is the successor of an even number, then we apply the induction hypothesis to give a number

`m`

and evidence that`m * 2 ≡ n`

. We return a pair consisting of`m`

and evidence that`1 + m * 2 ≡ suc n`

, which is immediate after substituting for`n`

.

This completes the proof in the forward direction.

Here is the proof in the reverse direction:∃-even : ∀ {n : ℕ} → ∃[ m ] ( m * 2 ≡ n) → even n ∃-odd : ∀ {n : ℕ} → ∃[ m ] (1 + m * 2 ≡ n) → odd n ∃-even ⟨ zero , refl ⟩ = even-zero ∃-even ⟨ suc m , refl ⟩ = even-suc (∃-odd ⟨ m , refl ⟩) ∃-odd ⟨ m , refl ⟩ = odd-suc (∃-even ⟨ m , refl ⟩)

Given a number that is twice some other number we must show it is even, and a number that is one more than twice some other number we must show it is odd. We induct over the evidence of the existential, and in the even case consider the two possibilities for the number that is doubled:

In the even case for

`zero`

, we must show`zero * 2`

is even, which follows by`even-zero`

.In the even case for

`suc n`

, we must show`suc m * 2`

is even. The inductive hypothesis tells us that`1 + m * 2`

is odd, from which the desired result follows by`even-suc`

.In the odd case, we must show

`1 + m * 2`

is odd. The inductive hypothesis tell us that`m * 2`

is even, from which the desired result follows by`odd-suc`

.

This completes the proof in the backward direction.

#### Exercise `∃-even-odd`

(practice)

How do the proofs become more difficult if we replace `m * 2`

and `1 + m * 2`

by `2 * m`

and `2 * m + 1`

? Rewrite the proofs of `∃-even`

and `∃-odd`

when restated in this way.

-- Your code goes here

#### Exercise `∃-+-≤`

(practice)

Show that `y ≤ z`

holds if and only if there exists a `x`

such that `x + y ≡ z`

.

-- Your code goes here

## Existentials, Universals, and Negation

Negation of an existential is isomorphic to the universal of a negation. Considering that existentials are generalised disjunction and universals are generalised conjunction, this result is analogous to the one which tells us that negation of a disjunction is isomorphic to a conjunction of negations:¬∃≃∀¬ : ∀ {A : Set} {B : A → Set} → (¬ ∃[ x ] B x) ≃ ∀ x → ¬ B x ¬∃≃∀¬ = record { to = λ{ ¬∃xy x y → ¬∃xy ⟨ x , y ⟩ } ; from = λ{ ∀¬xy ⟨ x , y ⟩ → ∀¬xy x y } ; from∘to = λ{ ¬∃xy → extensionality λ{ ⟨ x , y ⟩ → refl } } ; to∘from = λ{ ∀¬xy → refl } }

In the `to`

direction, we are given a value `¬∃xy`

of type `¬ ∃[ x ] B x`

, and need to show that given a value `x`

that `¬ B x`

follows, in other words, from a value `y`

of type `B x`

we can derive false. Combining `x`

and `y`

gives us a value `⟨ x , y ⟩`

of type `∃[ x ] B x`

, and applying `¬∃xy`

to that yields a contradiction.

In the `from`

direction, we are given a value `∀¬xy`

of type `∀ x → ¬ B x`

, and need to show that from a value `⟨ x , y ⟩`

of type `∃[ x ] B x`

we can derive false. Applying `∀¬xy`

to `x`

gives a value of type `¬ B x`

, and applying that to `y`

yields a contradiction.

The two inverse proofs are straightforward, where one direction requires extensionality.

#### Exercise `∃¬-implies-¬∀`

(recommended)

Show that existential of a negation implies negation of a universal:postulate ∃¬-implies-¬∀ : ∀ {A : Set} {B : A → Set} → ∃[ x ] (¬ B x) -------------- → ¬ (∀ x → B x)

Does the converse hold? If so, prove; if not, explain why.

#### Exercise `Bin-isomorphism`

(stretch)

Recall that Exercises Bin, Bin-laws, and Bin-predicates define a datatype `Bin`

of bitstrings representing natural numbers, and asks you to define the following functions and predicates:

```
to : ℕ → Bin
from : Bin → ℕ
Can : Bin → Set
```

And to establish the following properties:

```
from (to n) ≡ n
----------
Can (to n)
Can b
---------------
to (from b) ≡ b
```

Using the above, establish that there is an isomorphism between `ℕ`

and `∃[ b ] Can b`

.

We recommend proving the following lemmas which show that, for a given binary number `b`

, there is only one proof of `One b`

and similarly for `Can b`

.

```
≡One : ∀ {b : Bin} (o o′ : One b) → o ≡ o′
≡Can : ∀ {b : Bin} (c c′ : Can b) → c ≡ c′
```

Many of the alternatives for proving `to∘from`

turn out to be tricky. However, the proof can be straightforward if you use the following lemma, which is a corollary of `≡Can`

.

`proj₁≡→Can≡ : {c c′ : ∃[ b ] Can b} → proj₁ c ≡ proj₁ c′ → c ≡ c′`

-- Your code goes here

## Standard library

Definitions similar to those in this chapter can be found in the standard library:import Data.Product using (Σ; _,_; ∃; Σ-syntax; ∃-syntax)

## Unicode

This chapter uses the following unicode:

```
Π U+03A0 GREEK CAPITAL LETTER PI (\Pi)
Σ U+03A3 GREEK CAPITAL LETTER SIGMA (\Sigma)
∃ U+2203 THERE EXISTS (\ex, \exists)
```