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fix to even-odd in Quantifiers

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wadler 2018-03-20 13:47:39 -03:00
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2
src/Connectives.lagda
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@ -536,7 +536,7 @@ Right identity follows from commutativity of sum and left identity.
≃-∎
\end{code}
## Implication is function
## Implication is function {#implication}
Given two propositions `A` and `B`, the implication `A → B` holds if
whenever `A` holds then `B` must also hold. We formalise implication using

9
src/Preface.lagda
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@ -74,3 +74,12 @@ Most of the text was written during a sabbatical in the first half of 2018.
[stump]: http://www.morganclaypoolpublishers.com/catalog_Orig/product_info.php?cPath=24&products_id=908
[wen]: https://github.com/wenkokke
[phil]: http://homepages.inf.ed.ac.uk/wadler/
## A word on the exercises
Each exercise is followed by a name, which is the name you should use when
preparing an Agda file that solves the exercise. Sometimes it is up to you to
work out the type of the identifier, but sometimes we give it in the exercise.
In the latter case, the type is bound to an identifier with a capital in its
name, where the identifier you are to define has a small letter instead.

143
src/Quantifiers.lagda
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@ -213,11 +213,31 @@ Indeed, the converse also holds, and the two together form an isomorphism.
; to∘from = λ{ g → extensionality λ{ (x , y) → refl }}
}
\end{code}
The result can be viewed as a generalisation of currying, and the code to
establish the isomorphism is identical.
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][implication].
[implication]: Connectives/index.html#implication
### 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.
## Existential examples
## An existential example
Recall the definitions of `even` and `odd` from Chapter [Relations](Relations).
\begin{code}
@ -238,48 +258,28 @@ 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 ] ( 2 * m ≡ n)
odd n iff ∃[ m ] (1 + 2 * m ≡ n)
even n iff ∃[ m ] ( m * 2 ≡ n)
odd n iff ∃[ m ] (1 + m * 2 ≡ n)
First, we need a lemma, which states that twice the successor of `m`
is two more than twice `m`.
\begin{code}
lemma : ∀ (m : ℕ) → 2 * suc m ≡ 2 + 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)))
≡⟨⟩
2 + 2 * m
∎
\end{code}
The lemma is straightforward, and uses the lemma
`+-suc` from Chapter [Properties](Properties), which
allows us to simplify `m + suc n` to `suc (m + 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.
\begin{code}
even-∃ : ∀ {n : ℕ} → even n → ∃[ m ] ( 2 * m ≡ n)
odd-∃ : ∀ {n : ℕ} → odd n → ∃[ m ] (1 + 2 * m ≡ n)
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 , lemma m
... | m , refl = suc m , refl
odd-∃ (odd-suc e) with even-∃ e
... | m , refl = m , refl
\end{code}
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`?)
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
@ -287,48 +287,56 @@ consisting of zero and the (trivial) proof 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 + 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.
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 `2 * m ≡ n`. We return a pair consisting of `suc m`
and evidence that `1 + 2 * m = 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 `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}
∃-even : ∀ {n : ℕ} → ∃[ m ] ( 2 * m ≡ n) → even n
∃-odd : ∀ {n : ℕ} → ∃[ m ] (1 + 2 * m ≡ n) → odd n
∃-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) rewrite lemma m = even-suc (∃-odd (m , refl))
∃-even ( zero , refl) = even-zero
∃-even (suc m , refl) = even-suc (∃-odd (m , refl))
∃-odd ( m , refl) = odd-suc (∃-even (m , refl))
∃-odd ( m , refl) = odd-suc (∃-even (m , refl))
\end{code}
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.
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, if it is `zero`, then we must show `2 * zero` is
even, which follows by `even-zero`.
- In the even case for `zero`, we must show `2 * zero` is even, which
follows by `even-zero`.
- 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 even case for `suc n`, we must show `2 * suc 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`.
- 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′`)
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.
## Existentials, Universals, and Negation
Negation of an existential is isomorphic to universal
of a negation. Considering that existentials are generalised
disjunction and universals are generalised conjunction, this
@ -362,23 +370,6 @@ requires extensionality.
### 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)`.

31
src/extra/EvenOdd3.agda
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@ -0,0 +1,31 @@
-- Nils' suggestion
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 Relation.Nullary using (¬_)
open import Function using (_∘_; id)
open import Data.Product using (_×_; _,_; proj₁; proj₂; map; ∃; ∃-syntax)
open import Data.Sum using (_⊎_; inj₁; inj₂)
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)
∃-even : ∀ {n : ℕ} → even n → ∃[ m ] (n ≡ m * 2)
∃-odd : ∀ {n : ℕ} → odd n → ∃[ m ] (n ≡ 1 + m * 2)
∃-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