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) renaming (_,_ to ⟨_,_⟩) open import Data.Sum using (_⊎_) open import plfa.part1.Isomorphism using (_≃_; extensionality)
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.
Put another way, if we know that ∀ (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.
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 postulate a version of extensionality that
works for dependent functions.
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
.
We formalise existential quantification by declaring a suitable inductive type:
data Σ (A : Set) (B : A → Set) : Set where ⟨_,_⟩ : (x : A) → B x → Σ A B
We define a convenient syntax for existentials as follows:
Σsyntax = Σ infix 2 Σsyntax syntax Σsyntax A (λ x → B) = Σ[ x ∈ A ] B
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
Σsyntax
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.
Equivalently, we could also declare existentials as a record type:
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.
Given evidence that ∀ 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
.
Indeed, the converse also holds, and the two together form an isomorphism:
∀∃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 evenzero : even zero evensuc : ∀ {n : ℕ} → odd n  → even (suc n) data odd where oddsuc : ∀ {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∃ evenzero = ⟨ zero , refl ⟩ even∃ (evensuc o) with odd∃ o ...  ⟨ m , refl ⟩ = ⟨ suc m , refl ⟩ odd∃ (oddsuc 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 that1 + m * 2 ≡ n
. We return a pair consisting ofsuc m
and evidence thatsuc m * 2 ≡ suc n
, which is immediate after substituting forn
. 
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 thatm * 2 ≡ n
. We return a pair consisting ofsuc m
and evidence that1 + m * 2 ≡ suc n
, which is immediate after substituting forn
.
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 ⟩ = evenzero ∃even ⟨ suc m , refl ⟩ = evensuc (∃odd ⟨ m , refl ⟩) ∃odd ⟨ m , refl ⟩ = oddsuc (∃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 showzero * 2
is even, which follows byevenzero
. 
In the even case for
suc n
, we must showsuc m * 2
is even. The inductive hypothesis tells us that1 + m * 2
is odd, from which the desired result follows byevensuc
. 
In the odd case, we must show
1 + m * 2
is odd. The inductive hypothesis tell us thatm * 2
is even, from which the desired result follows byoddsuc
.
This completes the proof in the backward direction.
Exercise ∃evenodd
(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 Binisomorphism
(stretch)
Recall that Exercises
Bin,
Binlaws, and
Binpredicates
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 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} (cb : Can b) (cb' : Can b) → cb ≡ cb'
The proof of to∘from
is tricky. We recommend proving the following lemma
to∘fromaux : ∀ (b : Bin) (cb : Can b) → to (from b) ≡ b
→ _≡_ {_} {∃[ b ](Can b)} ⟨ to (from b) , canonto (from b) ⟩ ⟨ b , cb ⟩
You cannot immediately use ≡Can
to equate canonto (from b)
and
cb
because they have different types: Can (to (from b))
and Can b
respectively. You must first get their types to be equal, which
can be done by changing the type of cb
using rewrite
.
 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; _,_; ∃; Σsyntax; ∃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)