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further progress on Logic

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wadler 2018-01-26 19:35:55 -02:00
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@ -1171,6 +1171,23 @@ instantiate that proof that `∀ (x : A) → B[x] → C` to any `x`, and we
may choose the particular `x` provided by the evidence that `∃ (λ (x : may choose the particular `x` provided by the evidence that `∃ (λ (x :
A) → B[x])`. A) → B[x])`.
The types `¬ (∃ (λ (x : A) → B[x]))` and `∀ (x : A) → ¬ B[x]` are isomorphic.
\begin{code}
extensionality2 : ∀ {A B C : Set} → {f g : A → B → C} → (∀ (x : A) (y : B) → f x y ≡ g x y) → f ≡ g
extensionality2 fxy≡gxy = extensionality (λ x → extensionality (λ y → fxy≡gxy x y))
¬∃∀ : ∀ {A : Set} {B : A → Set} → (¬ ∃ (λ (x : A) → B x)) ≃ ∀ (x : A) → ¬ B x
¬∃∀ =
record
{ to = λ { ¬∃bx x bx → ¬∃bx (x , bx) }
; fro = λ { ∀¬bx (x , bx) → ∀¬bx x bx }
; invˡ = λ { ¬∃bx → extensionality (λ { (x , bx) → refl }) }
; invʳ = λ { ∀¬bx → refl }
}
\end{code}
[It would be better to have even and odd as an exercise. Is there [It would be better to have even and odd as an exercise. Is there
a simpler example that I could start with?] a simpler example that I could start with?]
@ -1188,10 +1205,6 @@ We show that a number `n` is even if and only if there exists
another number `m` such that `n ≡ 2 * m`, and is odd if and only another number `m` such that `n ≡ 2 * m`, and is odd if and only
if there is another number `m` such that `n ≡ 1 + 2 * m`. if there is another number `m` such that `n ≡ 1 + 2 * m`.
Here is the proof in the forward direction.
\begin{code}
\end{code}
## Decidability ## Decidability