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completed pass over Quantifiers

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wadler 2018-03-18 12:27:39 -03:00
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60
src/Connectives.lagda
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@ -694,7 +694,7 @@ is the same as the assertion that if `A` holds then `B` holds and if
`A` holds then `C` holds. The proof of left inverse requires both extensionality
and the rule `η-×` for products.
\begin{code}
→-distrib-× : ∀ {A B C : Set} → (A → B × C) ≃ ((A → B) × (A → C))
→-distrib-× : ∀ {A B C : Set} → (A → B × C) ≃ (A → B) × (A → C)
→-distrib-× =
record
{ to = λ{ f → ( (λ{ x → proj₁ (f x) }) , (λ{ y → proj₂ (f y)}) ) }
@ -710,39 +710,39 @@ and the rule `η-×` for products.
Products distributes over sum, up to isomorphism. The code to validate
this fact is similar in structure to our previous results.
\begin{code}
×-distrib-⊎ : ∀ {A B C : Set} → ((A ⊎ B) × C) ≃ ((A × C) ⊎ (B × C))
×-distrib-⊎ : ∀ {A B C : Set} → (A ⊎ B) × C ≃ (A × C) ⊎ (B × C)
×-distrib-⊎ =
record
{ to = λ { ((inj₁ x) , z) → (inj₁ (x , z))
; ((inj₂ y) , z) → (inj₂ (y , z))
}
; from = λ { (inj₁ (x , z)) → ((inj₁ x) , z)
; (inj₂ (y , z)) → ((inj₂ y) , z)
}
; from∘to = λ { ((inj₁ x) , z) → refl
; ((inj₂ y) , z) → refl
}
; to∘from = λ { (inj₁ (x , z)) → refl
; (inj₂ (y , z)) → refl
}
{ to = λ{ ((inj₁ x) , z) → (inj₁ (x , z))
; ((inj₂ y) , z) → (inj₂ (y , z))
}
; from = λ{ (inj₁ (x , z)) → ((inj₁ x) , z)
; (inj₂ (y , z)) → ((inj₂ y) , z)
}
; from∘to = λ{ ((inj₁ x) , z) → refl
; ((inj₂ y) , z) → refl
}
; to∘from = λ{ (inj₁ (x , z)) → refl
; (inj₂ (y , z)) → refl
}
}
\end{code}
Sums do not distribute over products up to isomorphism, but it is an embedding.
\begin{code}
⊎-distrib-× : ∀ {A B C : Set} → ((A × B) ⊎ C) ≲ ((A ⊎ C) × (B ⊎ C))
⊎-distrib-× : ∀ {A B C : Set} → (A × B) ⊎ C ≲ (A ⊎ C) × (B ⊎ C)
⊎-distrib-× =
record
{ to = λ { (inj₁ (x , y)) → (inj₁ x , inj₁ y)
; (inj₂ z) → (inj₂ z , inj₂ z)
}
; from = λ { (inj₁ x , inj₁ y) → (inj₁ (x , y))
; (inj₁ x , inj₂ z) → (inj₂ z)
; (inj₂ z , _ ) → (inj₂ z)
}
; from∘to = λ { (inj₁ (x , y)) → refl
; (inj₂ z) → refl
}
{ to = λ{ (inj₁ (x , y)) → (inj₁ x , inj₁ y)
; (inj₂ z) → (inj₂ z , inj₂ z)
}
; from = λ{ (inj₁ x , inj₁ y) → (inj₁ (x , y))
; (inj₁ x , inj₂ z) → (inj₂ z)
; (inj₂ z , _ ) → (inj₂ z)
}
; from∘to = λ{ (inj₁ (x , y)) → refl
; (inj₂ z) → refl
}
}
\end{code}
Note that there is a choice in how we write the `from` function.
@ -763,6 +763,16 @@ gives rise to an isomorphism, while the second only gives rise to an
embedding, revealing a sense in which one of these laws is "more
true" than the other.
### Exercise (`×⊎-implies-⊎×`)
Show that a conjunct of disjuncts implies a disjunct of conjuncts.
\begin{code}
×⊎-Implies-⊎× = ∀ {A B C D : Set} → (A ⊎ B) × (C ⊎ D) → (A × C) ⊎ (B × D)
\end{code}
Does the converse hold? If so, prove; if not, explain why.
### Exercise (`⇔-refl`, `⇔-sym`, `⇔-trans`)
Define equivalence of propositions (also known as "if and only if") as follows.

2
src/Isomorphism.lagda
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@ -35,6 +35,7 @@ then applies `g`.
In set theory, two sets are isomorphic if they are in one-to-one correspondence.
Here is a formal definition of isomorphism.
\begin{code}
infix 0 _≃_
record _≃_ (A B : Set) : Set where
field
to : A → B
@ -216,6 +217,7 @@ correspondence between the second type and the first.
Here is the formal definition of embedding.
\begin{code}
infix 0 _≲_
record _≲_ (A B : Set) : Set where
field
to : A → B

175
src/Quantifiers.lagda
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@ -93,7 +93,7 @@ Show that a disjunction of universals implies a universal of disjunctions.
⊎∀-Implies-∀⊎ = ∀ {A : Set} { B C : A → Set } →
(∀ (x : A) → B x) ⊎ (∀ (x : A) → C x) → ∀ (x : A) → B x ⊎ C x
\end{code}
Does the converse also hold? If so, prove; if not, explain why.
Does the converse hold? If so, prove; if not, explain why.
## Existentials
@ -133,24 +133,29 @@ are provided by the evidence for `∃ (λ (x : A) → B x)`.
Agda makes it possible to define our own syntactic abbreviations.
\begin{code}
syntax ∃ {A} (λ x → B) = ∃[ x ∈ A ] B
syntax ∃ (λ x → B) = ∃[ x ] B
\end{code}
This allows us to write `∃[ x ∈ A ] (B x)` in place of
`∃ (λ (x : A) → B x)`.
This allows us to write `∃[ x ] (B x)` in place of `∃ (λ x → B x)`.
As an example, recall the definitions of `even` from
As an example, recall the definitions of `even` and `odd` from
Chapter [Relations](Relations).
\begin{code}
data even : ℕ → Set where
ev0 : even zero
ev+2 : ∀ {n : ℕ} → even n → even (suc (suc n))
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)
\end{code}
A number is even if it is zero, or if it is two
greater than an even number.
A number is even if it is zero or the successor of an odd number, and
odd if it the successor of an even number.
We will show that a number is even if and only if it is twice some
other number. That is, number `n` is even if and only if there exists
a number `m` such that twice `m` is `n`.
other number, and odd if and only if it is one more than twice
some other number.
First, we need a lemma, which allows us to
simplify twice the successor of `m` to two
@ -175,48 +180,71 @@ The lemma is straightforward, and uses the lemma
allows us to simplify `m + suc n` to `suc (m + n)`.
Here is the proof in the forward direction.
\begin{code}
ev-ex : ∀ {n : ℕ} → even n → ∃[ m ∈ ℕ ] (2 * m ≡ n)
ev-ex ev0 = (zero , refl)
ev-ex (ev+2 ev) with ev-ex ev
... | (m , refl) = (suc m , lemma m)
\begin{code}
even-∃ : ∀ {n : ℕ} → even n → ∃[ m ] ( 2 * m ≡ n)
odd-∃ : ∀ {n : ℕ} → odd n → ∃[ m ] (1 + 2 * m ≡ n)
even-∃ even-zero = zero , refl
even-∃ (even-suc o) with odd-∃ o
... | m , refl = suc m , lemma m
odd-∃ (odd-suc e) with even-∃ e
... | m , refl = m , refl
\end{code}
Given an even number, we must show it is twice some
other number. The proof is a function, which
given a proof that `n` is even
returns a pair consisting of `m` and a proof
that twice `m` is `n`. The proof is by induction over
the evidence that `n` is even.
We define two mutually recursive functions. Given
evidence that `n` is even or odd, we return a
number `m` and evidence that `2 * m ≡ n` or `1 + 2 * m ≡ n`.
(By convention, one tends to put a constant at the
end of a term, so why have we chosen here
to write `1 + 2 * m` rather than `2 * m + 1`?)
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 (trivial) proof that twice zero is zero.
- If the number is even because it is two more than another even
number `n`, then we apply the induction hypothesis, giving us a number
`m` and a proof that `2 * m ≡ n`, which we match against `refl`. We
return a pair consisting of `suc m` and a proof that `2 * suc m ≡ suc
(suc n)`, which follows from `2 * m ≡ n` and the lemma.
- 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 + 2 * m ≡ n`. We return a pair consisting of `suc m`
and evidence that `2 * (suc m)` ≡ suc n`, which after substituting for
`n` means we need to show `2 * (suc m) ≡ 2 + 2 * m`, which follows
from our lemma.
- 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 `2 * m ≡ n`. We return a pair conisting of `suc m`
and evidence that `1 + 2 * m = 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.
\begin{code}
ex-ev : ∀ {n : ℕ} → ∃[ m ∈ ℕ ] (2 * m ≡ n) → even n
ex-ev (zero , refl) = ev0
ex-ev (suc m , refl) rewrite lemma m = ev+2 (ex-ev (m , refl))
∃-even : ∀ {n : ℕ} → ∃[ m ] ( 2 * m ≡ n) → even n
∃-odd : ∀ {n : ℕ} → ∃[ m ] (1 + 2 * m ≡ n) → odd n
∃-even ( zero , refl) = even-zero
∃-even (suc m , refl) rewrite lemma m = even-suc (∃-odd (m , refl))
∃-odd ( m , refl) = odd-suc (∃-even (m , refl))
\end{code}
Given a number that is twice some other number,
we must show that it is even. The proof is a function,
which given a number `m` and a proof that `n` is twice `m`,
returns a proof that `n` is even. The proof is by induction
over the number `m`.
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 particular the number that is doubled.
- If it is zero, then we must show that twice zero is even,
which follows by rule `ev0`.
- In the even case, if it is `zero`, then we must show `2 * zero` is
even, which follows by `even-zero`.
- If it is `suc m`, then we must show that `2 * suc m` is
even. After rewriting with our lemma, we must show that
`suc (suc (2 * m))` is even. The inductive hypothesis tells
us `2 * m` is even, from which the desired result follows
by rule `ev+2`.
- In the even case, if it is `suc n`, then we must show `2 * suc m` is
even. After rewriting with our lemma, we must show that `2 + 2 * m`
is even. The inductive hypothesis tells us that `1 + 2 * m` is odd,
from which the desired result follows by `even-suc`.
- In the odd case, then we must show `1 + 2 * suc m` is odd. The
inductive hypothesis tell us that `2 * m` is even, from which the
desired result follows by `odd-suc`.
This completes the proof in the backward direction.
Negation of an existential is isomorphic to universal
of a negation. Considering that existentials are generalised
@ -224,44 +252,67 @@ disjuntion and universals are generalised conjunction, this
result is analogous to the one which tells us that negation
of a disjuntion is isomorphic to a conjunction of negations.
\begin{code}
¬∃∀ : ∀ {A : Set} {B : A → Set} → (¬ ∃[ x ∈ A ] B x) ≃ ∀ (x : A) → ¬ B x
¬∃∀ : ∀ {A : Set} {B : A → Set} → (¬ ∃[ x ] B x) ≃ ∀ x → ¬ B x
¬∃∀ =
record
{ to = λ { ¬∃bx x bx → ¬∃bx (x , bx) }
; from = λ { ∀¬bx (x , bx) → ∀¬bx x bx }
; from∘to = λ { ¬∃bx → extensionality (λ { (x , bx) → refl }) }
; to∘from = λ { ∀¬bx → refl }
{ 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 }
}
\end{code}
In the `to` direction, we are given a value `¬∃bx` of type
`¬ ∃ (λ (x : A) → B x)`, and need to show that given a value
`x` of type `A` that `¬ B x` follows, in other words, from
a value `bx` of type `B x` we can derive false. Combining
`x` and `bx` gives us a value `(x , bx)` of type `∃ (λ (x : A) → B x)`,
and applying `¬∃bx` to that yields a contradiction.
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 `∀¬bx` of type
`∀ (x : A) → ¬ B x`, and need to show that from a value `(x , bx)`
of type `∃ (λ (x : A) → B x)` we can derive false. Applying `∀¬bx`
to `x` gives a value of type `¬ B x`, and applying that to `bx` yields
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* Show `∃ (λ (x : A) → ¬ B x) → ¬ (∀ (x : A) → B x)`.
### Exercise (`∃-distrib-⊎`)
Show that universals distribute over conjunction.
\begin{code}
∃-Distrib-⊎ = ∀ {A : Set} {B C : A → Set} →
∃[ x ] (B x ⊎ C x) ≃ (∃[ x ] B x) ⊎ (∃[ x ] C x)
\end{code}
### Exercise (`∃×-implies-×∃`)
Show that an existential of conjunctions implies a conjunction of existentials.
\begin{code}
∃×-Implies-×∃ = ∀ {A : Set} { B C : A → Set } →
∃[ x ] (B x × C x) → (∃[ x ] B x) × (∃[ x ] C x)
\end{code}
Does the converse hold? If so, prove; if not, explain why.
### Exercise (`∃¬-Implies-¬∀)
Show `∃[ x ] ¬ B x → ¬ (∀ x → B x)`.
Does the converse hold? If so, prove; if not, explain why.
## Standard Prelude
Definitions similar to those in this chapter can be found in the standard library.
\begin{code}
import Data.Product using (∃)
import Data.Product using (∃;_,_)
\end{code}
## Unicode
This chapter introduces the following unicode.
This chapter uses the following unicode.
≤ U+2264 LESS-THAN OR EQUAL TO (\<=, \le)
∃ U+2203 THERE EXISTS (\ex, \exists)

2
src/Relations.lagda
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@ -540,7 +540,7 @@ data odd where
odd-suc : ∀ {n : ℕ} → even n → odd (suc n)
\end{code}
A number is even if it is zero or the successor of an odd number,
and a number is odd if it is the successor of an even number.
and odd if it is the successor of an even number.
This is our first use of a mutually recursive datatype declaration.
Since each identifier must be defined before it is used, we first

18
src/extra/DecidableBroken.agda
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@ -14,22 +14,16 @@ suc m ≤? suc n with m ≤? n
... | yes m≤n = yes (s≤s m≤n)
... | no ¬m≤n = no λ{ (s≤s m≤n) → ¬m≤n m≤n }
_ : Dec (2 ≤ 4)
_ = 2 ≤? 4 -- yes (s≤s (s≤s z≤n))
_ : 2 ≤? 4 ≡ yes (s≤s (s≤s z≤n))
_ = refl
_ : Dec (4 ≤ 2)
_ = 4 ≤? 2 -- no λ{(s≤s (s≤s ()))}
_ : 4 ≤? 2 ≡ no (λ { (s≤s m≤n) → (λ { (s≤s m≤n) → (λ ()) m≤n }) m≤n })
_ : 4 ≤? 2 ≡ no {!!}
_ = refl
{-
/Users/wadler/sf/src/extra/DecidableBroken.agda:27,5-9
(λ { (s≤s m≤n) → (λ { (s≤s m≤n) → (λ ()) 1 m≤n }) m≤n }) x !=
(λ { (s≤s m≤n) → _34 m≤n }) x of type .Data.Empty.⊥
when checking that the expression refl has type
(4 ≤? 2) ≡ no (λ { (s≤s m≤n) → _34 m≤n })
Using ^C ^N, the term
4 ≤? 2
evaluates to
no (λ { (s≤s m≤n) → (λ { (s≤s m≤n) → (λ ()) 1 m≤n }) m≤n })
The 1 is spurious.
-}

34
src/extra/RewriteWTF.agda
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@ -1,5 +1,6 @@
import Relation.Binary.PropositionalEquality as Eq
open Eq using (_≡_; refl; sym; trans; cong)
open Eq.≡-Reasoning
open import Data.Nat using (ℕ; zero; suc; _+_; _*_)
open import Data.Nat.Properties.Simple using (+-suc)
open import Data.Product using (∃; ∃-syntax; _,_)
@ -14,23 +15,46 @@ data even where
data odd where
odd-suc : ∀ {n : ℕ} → even n → odd (suc n)
∃-even : ∀ {n : ℕ} → even n → ∃[ m ] (n ≡ 2 * m)
∃-odd : ∀ {n : ℕ} → odd n → ∃[ m ] (n ≡ 1 + 2 * m)
lemma : ∀ (m : ℕ) → 2 * suc m ≡ suc (suc (2 * m))
lemma m =
begin
2 * suc m
≡⟨⟩
suc m + (suc m + zero)
≡⟨⟩
suc (m + (suc (m + zero)))
≡⟨ cong suc (+-suc m (m + zero)) ⟩
suc (suc (m + (m + zero)))
≡⟨⟩
suc (suc (2 * m))
∎
∃-even : ∀ {n : ℕ} → even n → ∃[ m ] (2 * m ≡ n)
∃-odd : ∀ {n : ℕ} → odd n → ∃[ m ] (1 + 2 * m ≡ n)
∃-even even-zero = zero , refl
∃-even (even-suc o) with ∃-odd o
... | m , eqn rewrite eqn | sym (+-suc m (m + 0)) = suc m , refl
... | m , refl = suc m , lemma m
∃-odd (odd-suc e) with ∃-even e
... | m , eqn rewrite eqn = m , refl
... | m , refl = m , refl
∃-even′ : ∀ {n : ℕ} → even n → ∃[ m ] (n ≡ 2 * m)
∃-odd′ : ∀ {n : ℕ} → odd n → ∃[ m ] (n ≡ 1 + 2 * m)
∃-even′ even-zero = zero , refl
∃-even′ (even-suc o) with ∃-odd′ o
... | m , eqn rewrite eqn | +-suc m (m + 0) = suc m , ?
... | m , eqn rewrite eqn | +-suc m (m + 0) = suc m , {!!}
∃-odd′ (odd-suc e) with ∃-even′ e
... | m , eqn rewrite eqn = m , refl
data Even : ℕ → Set where
ev0 : Even zero
ev2 : ∀ {n} → Even n → Even (suc (suc n))
ev-ex : ∀ {n : ℕ} → Even n → ∃[ m ] (2 * m ≡ n)
ev-ex ev0 = (zero , refl)
ev-ex (ev2 ev) with ev-ex ev
... | (m , refl) = (suc m , lemma m)