```module Assignment2 where
```

## Introduction

You must do all the exercises labelled “(recommended)”.

Exercises labelled “(stretch)” are there to provide an extra challenge. You don’t need to do all of these, but should attempt at least a few.

Exercises labelled “(practice)” are included for those who want extra practice.

Submit your homework using Gradescope. Go to the course page under Learn. Select `Assessment` from the left hand menu, then select `Assignment Submission`.

## Good Scholarly Practice.

Furthermore, you are required to take reasonable measures to protect your assessed work from unauthorised access. For example, if you put any such work on a public repository then you must set access permissions appropriately (generally permitting access only to yourself, or your group in the case of group practicals).

## Equality

```module Equality where
```

## Imports

This chapter has no imports. Every chapter in this book, and nearly every module in the Agda standard library, imports equality. Since we define equality here, any import would create a conflict.

```  import Relation.Binary.PropositionalEquality as Eq
open Eq using (_≡_; refl; cong; cong-app)
open Eq.≡-Reasoning
```

#### Exercise `trans` and `≡-Reasoning` (practice)

Sadly, we cannot use the definition of trans’ using ≡-Reasoning as the definition for trans. Can you see why? (Hint: look at the definition of `_≡⟨_⟩_`)

```  -- Your code goes here
```

#### Exercise `≤-Reasoning` (stretch)

The proof of monotonicity from Chapter Relations can be written in a more readable form by using an analogue of our notation for `≡-Reasoning`. Define `≤-Reasoning` analogously, and use it to write out an alternative proof that addition is monotonic with regard to inequality. Rewrite all of `+-monoˡ-≤`, `+-monoʳ-≤`, and `+-mono-≤`.

```  -- Your code goes here
```

## Isomorphism

```module Isomorphism where
```

## Imports

```  import Relation.Binary.PropositionalEquality as Eq
open Eq using (_≡_; refl; cong; cong-app)
open Eq.≡-Reasoning
open import Data.Nat using (ℕ; zero; suc; _+_)
open import Data.Nat.Properties using (+-comm)
```
```  open import plfa.part1.Isomorphism
hiding (≃-implies-≲; _⇔_)
```

#### Exercise `≃-implies-≲` (practice)

Show that every isomorphism implies an embedding.
```  postulate
≃-implies-≲ : ∀ {A B : Set}
→ A ≃ B
-----
→ A ≲ B
```
```  -- Your code goes here
```

#### Exercise `_⇔_` (practice)

Define equivalence of propositions (also known as “if and only if”) as follows:
```  record _⇔_ (A B : Set) : Set where
field
to   : A → B
from : B → A
```

Show that equivalence is reflexive, symmetric, and transitive.

```  -- Your code goes here
```

#### Exercise `Bin-embedding` (stretch)

Recall that Exercises Bin and Bin-laws 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``````

which satisfy the following property:

``from (to n) ≡ n``
Using the above, establish that there is an embedding of `ℕ` into `Bin`.
```  -- Your code goes here
```

Why do `to` and `from` not form an isomorphism?

## Connectives

```module Connectives where
```

## Imports

```  import Relation.Binary.PropositionalEquality as Eq
open Eq using (_≡_; refl)
open Eq.≡-Reasoning
open import Data.Nat using (ℕ)
open import Function using (_∘_)
open import plfa.part1.Isomorphism using (_≃_; _≲_; extensionality)
open plfa.part1.Isomorphism.≃-Reasoning
```
```  open import plfa.part1.Connectives
hiding (⊎-weak-×; ⊎×-implies-×⊎)
```

Show that `A ⇔ B` as defined earlier is isomorphic to `(A → B) × (B → A)`.

```  -- Your code goes here
```

Show sum is commutative up to isomorphism.

```  -- Your code goes here
```

#### Exercise `⊎-assoc` (practice)

Show sum is associative up to isomorphism.

```  -- Your code goes here
```

Show empty is the left identity of sums up to isomorphism.

```  -- Your code goes here
```

#### Exercise `⊥-identityʳ` (practice)

Show empty is the right identity of sums up to isomorphism.

```  -- Your code goes here
```
Show that the following property holds:
```  postulate
⊎-weak-× : ∀ {A B C : Set} → (A ⊎ B) × C → A ⊎ (B × C)
```

This is called a weak distributive law. Give the corresponding distributive law, and explain how it relates to the weak version.

```  -- Your code goes here
```

#### Exercise `⊎×-implies-×⊎` (practice)

Show that a disjunct of conjuncts implies a conjunct of disjuncts:
```  postulate
⊎×-implies-×⊎ : ∀ {A B C D : Set} → (A × B) ⊎ (C × D) → (A ⊎ C) × (B ⊎ D)
```

Does the converse hold? If so, prove; if not, give a counterexample.

```  -- Your code goes here
```

## Negation

```module Negation where
```

## Imports

```  open import Relation.Binary.PropositionalEquality using (_≡_; refl)
open import Data.Nat using (ℕ; zero; suc)
open import Data.Empty using (⊥; ⊥-elim)
open import Data.Sum using (_⊎_; inj₁; inj₂)
open import Data.Product using (_×_)
open import plfa.part1.Isomorphism using (_≃_; extensionality)
```
```  open import plfa.part1.Negation
hiding (Stable)
```

Using negation, show that strict inequality is irreflexive, that is, `n < n` holds for no `n`.

```  -- Your code goes here
```

#### Exercise `trichotomy` (practice)

Show that strict inequality satisfies trichotomy, that is, for any naturals `m` and `n` exactly one of the following holds:

• `m < n`
• `m ≡ n`
• `m > n`

Here “exactly one” means that not only one of the three must hold, but that when one holds the negation of the other two must also hold.

```  -- Your code goes here
```

Show that conjunction, disjunction, and negation are related by a version of De Morgan’s Law.

``¬ (A ⊎ B) ≃ (¬ A) × (¬ B)``

This result is an easy consequence of something we’ve proved previously.

```  -- Your code goes here
```

Do we also have the following?

``¬ (A × B) ≃ (¬ A) ⊎ (¬ B)``

If so, prove; if not, can you give a relation weaker than isomorphism that relates the two sides?

#### Exercise `Classical` (stretch)

Consider the following principles:

• Excluded Middle: `A ⊎ ¬ A`, for all `A`
• Double Negation Elimination: `¬ ¬ A → A`, for all `A`
• Peirce’s Law: `((A → B) → A) → A`, for all `A` and `B`.
• Implication as disjunction: `(A → B) → ¬ A ⊎ B`, for all `A` and `B`.
• De Morgan: `¬ (¬ A × ¬ B) → A ⊎ B`, for all `A` and `B`.

Show that each of these implies all the others.

```  -- Your code goes here
```

#### Exercise `Stable` (stretch)

Say that a formula is stable if double negation elimination holds for it:
```  Stable : Set → Set
Stable A = ¬ ¬ A → A
```

Show that any negated formula is stable, and that the conjunction of two stable formulas is stable.

```  -- Your code goes here
```

## Quantifiers

```module Quantifiers where
```

## 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; proj₂) renaming (_,_ to ⟨_,_⟩)
open import Data.Sum using (_⊎_; inj₁; inj₂)
open import plfa.part1.Isomorphism using (_≃_; extensionality)
```
```  open import plfa.part1.Quantifiers
hiding (∀-distrib-×; ⊎∀-implies-∀⊎; ∃-distrib-⊎; ∃×-implies-×∃; ∃¬-implies-¬∀; Tri)
```
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.

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`.

#### Exercise `∃-even-odd` (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
```
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 `Bin-isomorphism` (stretch)

Recall that Exercises Bin, Bin-laws, and Bin-predicates 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 the 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 cb′ : Can b) → cb ≡ cb′``````

Many of the alternatives for proving `to∘from` turn out to be tricky. However, the proof can be straightforward if you use the following lemma, which is a corollary of `≡Can`.

``proj₁≡→Can≡ : {cb cb′ : ∃[ b ] Can b} → proj₁ cb ≡ proj₁ cb′ → cb ≡ cb′``
```  -- Your code goes here
```