Merge branch 'dev' of github.com:plfa/plfa.github.io into dev
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1 changed files with 5 additions and 5 deletions
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@ -105,7 +105,7 @@ Show that a disjunction of universals implies a universal of disjunctions:
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```
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postulate
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⊎∀-implies-∀⊎ : ∀ {A : Set} {B C : A → Set} →
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(∀ (x : A) → B x) ⊎ (∀ (x : A) → C x) → ∀ (x : A) → B x ⊎ C x
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(∀ (x : A) → B x) ⊎ (∀ (x : A) → C x) → ∀ (x : A) → B x ⊎ C x
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```
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Does the converse hold? If so, prove; if not, explain why.
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@ -463,21 +463,21 @@ And to establish the following properties:
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to (from b) ≡ b
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Using the above, establish that there is an isomorphism between `ℕ` and
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`∃[ b ](Can b)`.
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`∃[ b ] Can b`.
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We recommend proving the following lemmas which show that, for a given
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binary number `b`, there is only one proof of `One b` and similarly
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for `Can b`.
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≡One : ∀{b : Bin} (o o' : One b) → o ≡ o'
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≡One : ∀ {b : Bin} (o o′ : One b) → o ≡ o′
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≡Can : ∀{b : Bin} (cb : Can b) (cb' : Can b) → cb ≡ cb'
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≡Can : ∀ {b : Bin} (cb cb′ : Can b) → cb ≡ cb′
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Many of the alternatives for proving `to∘from` turn out to be tricky.
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However, the proof can be straightforward if you use the following lemma,
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which is a corollary of `≡Can`.
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proj₁≡→Can≡ : {cb cb′ : ∃[ b ](Can b)} → proj₁ cb ≡ proj₁ cb′ → cb ≡ cb′
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proj₁≡→Can≡ : {cb cb′ : ∃[ b ] Can b} → proj₁ cb ≡ proj₁ cb′ → cb ≡ cb′
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```
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-- Your code goes here
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