module Data.Int.HIT where

The integers as a HIT🔗

One of the possible characterisations of the integers is as the group freely generated by the monoid of natural numbers under addition. This module constructs this as a higher inductive type:

data Int : Type where
  diff : (x y : Nat) → Int
  quot : (m n : Nat) → diff m n ≡ diff (suc m) (suc n)
\ Warning

While this higher-inductive type is a perfectly workable definition of the integers from a mathematical point of view, the ergonomics of higher inductive types are still an area of active research. Therefore, we prefer a “boring” inductive representation of the integers.

This is an alternative representation of the construction of the group integers as consisting of pairs where is set identical to when The single generating path can be stretched to obtain this more familiar quotient, but it has the benefit of provably resulting in a set (without truncation). Intuitively, this is because we only identify points that “used to be distinct”.

We can start by proving that all representations of zero are the same:

zeroes : (n : Nat) → diff 0 0 ≡ diff n n
zeroes zero    = refl
zeroes (suc n) = zeroes n ∙ quot _ _

Additionally, offsetting a difference by a fixed natural, as long as it’s done on both sides of the difference, does not change which integer is being represented: That is, considering all three naturals as integers,

cancel : (a b n : Nat) → diff a b ≡ diff (n + a) (n + b)
cancel a b zero    = refl
cancel a b (suc n) = cancel a b n ∙ quot _ _

As a final pair of helper lemmas, we find that if and differ by an absolute value of then the values and are the same (as long as we fix the sign — hence the two lemmas). The generic situation of “differing by ” is captured by fixing a natural number and adding because we have and

offset-negative : (a b : Nat) → diff a (a + b) ≡ diff 0 b
offset-negative zero b = refl
offset-negative (suc a) b =
  diff (suc a) (suc (a + b)) ≡⟨ sym (quot _ _) ⟩≡
  diff a (a + b)             ≡⟨ offset-negative a b ⟩≡
  diff 0 b                   ∎

offset-positive : (a b : Nat) → diff (a + b) a ≡ diff b 0
offset-positive zero b = refl
offset-positive (suc a) b =
  diff (suc (a + b)) (suc a) ≡⟨ sym (quot _ _) ⟩≡
  diff (a + b) a             ≡⟨ offset-positive a b ⟩≡
  diff b 0                   ∎

Those two are the last two lemmas we need to prove the “if” direction of “naturals are identified in the quotient iff they represent the same difference”: the construction same-difference below packages everything together with a bow on the top.

opaque
  same-difference : {a b c d : Nat} → a + d ≡ b + c → diff a b ≡ diff c d
  same-difference {zero} {b} {c} {d} path = sym $
    diff c ⌜ d ⌝     ≡⟨ ap! path ⟩≡
    diff c ⌜ b + c ⌝ ≡⟨ ap! (+-commutative b c) ⟩≡
    diff c (c + b)   ≡⟨ offset-negative _ _ ⟩≡
    diff 0 b         ∎
  same-difference {suc a} {zero} {c} {d} path = sym $
    diff ⌜ c ⌝ d         ≡⟨ ap! (sym path) ⟩≡
    diff ⌜ suc a + d ⌝ d ≡⟨ ap! (+-commutative (suc a) d) ⟩≡
    diff (d + suc a) d   ≡⟨ offset-positive _ _ ⟩≡
    diff (suc a) 0       ∎

  same-difference {suc a} {suc b} {c} {d} path =
    diff (suc a) (suc b) ≡˘⟨ quot _ _ ⟩≡˘
    diff a b             ≡⟨ same-difference (suc-inj path) ⟩≡
    diff c d             ∎

In the other direction, we must be clever: we use path induction, defining a type family such that the fibre of over is We can then use path induction to construct the map inverse to same-difference. On the way, the first thing we establish is a pair of observations about equalities on the natural numbers: and are equivalent conditions. This can be seen by commutativity and injectivity of the successor function, but below we prove it using equational reasoning, without appealing to commutativity.

module ℤ-Path where
  private variable a b m n c d : Nat

  encode-p-from : (a + n ≡ b + m) → (a + suc n ≡ b + suc m)
  encode-p-from {a = a} {n} {b} {m} p =
    a + suc n   ≡⟨ +-sucr a n ⟩≡
    suc (a + n) ≡⟨ ap suc p ⟩≡
    suc (b + m) ≡˘⟨ +-sucr b m ⟩≡˘
    b + suc m   ∎

  encode-p-to : (a + suc n ≡ b + suc m) → (a + n ≡ b + m)
  encode-p-to {a} {n} {b} {m} p = suc-inj (sym (+-sucr a n) ∙ p ∙ +-sucr b m)

We then define, fixing two natural numbers the family by recursion on the integer Recall that we want the fibre over to be so that’s our pick. Now, the quot path constructor mandates that the fibre over be the same as that over — but this follows by propositional extensionality and the pair of observations above.

  Code : ∀ (a b : Nat) (x : Int) → Type
  Code a b (diff c d) = a + d ≡ b + c
  Code a b (quot m n i) = path i where
    path : (a + n ≡ b + m) ≡ (a + suc n ≡ b + suc m)
    path = ua (prop-ext (Nat-is-set _ _) (Nat-is-set _ _) encode-p-from encode-p-to)

Hence, if we have a path we can apply path induction, whence it suffices to consider the case where is literally_ the difference of and To lift this into our Code fibration, we must show that but this is exactly commutativity of addition on

  encode : ∀ (a b : Nat) (x : Int) → diff a b ≡ x → Code a b x
  encode a b x = J (λ x p → Code a b x) (+-commutative a b)

As a finishing touch, we give Int instances for Number and Negative, meaning that we can use positive and negative integer literals to denote values of Int.

instance
  Number-Int : Number Int
  Number-Int .Number.Constraint _ = ⊤
  Number-Int .Number.fromNat n = diff n 0

  Negative-Int : Negative Int
  Negative-Int .Negative.Constraint _ = ⊤
  Negative-Int .Negative.fromNeg n = diff 0 n

Canonical representatives🔗

Initially, we note that the type of integers admits a surjection from the type given by sending each pair of naturals to their difference.

private
  difference-surjection : ∀ x → ∃ (Nat × Nat) (λ (a , b) → (diff a b ≡ x))
  difference-surjection (diff x y) = inc ((x , y) , refl)
  difference-surjection (quot m n i) =
    is-prop→pathp
      (λ i → ∥_∥.squash {A = Σ[ (a , b) ∈ Nat × Nat ] (diff a b ≡ quot m n i)})
      (inc ((m , n) , refl))
      (inc ((suc m , suc n) , refl))
      i

What we’ll show is that this surjection actually splits: Given an integer, we can find out what natural numbers it came from. Well, not quite: We can find a reduced representation of that difference. Namely, suppose we’re given the integer We split by cases:

  • If then this is the same integer as
  • If then this is the same integer as
  • If then this is the same integer as

A “canonical form” for an integer is a pair of natural numbers that represent (under diff) the same integer we started with. The canonicalisation procedure does the split we described above, appealing to a battery of three lemmas to prove the equality.

Canonical : Int → Type
Canonical n = Σ[ x ∈ Nat ] Σ[ y ∈ Nat ] (diff x y ≡ n)

canonicalise : (n : Int) → Canonical n
canonicalise = go where
  lemma₁ : ∀ x y → .(x < y) → diff 0 (y - x) ≡ diff x y
  lemma₂ : ∀ x y → .(y < x) → diff (x - y) 0 ≡ diff x y
  lemma₃ : ∀ x y → .(x ≡ y) → diff 0 0       ≡ diff x y

  work : ∀ x y → Canonical (diff x y)
  work x y with ≤-split x y
  ... | inl p       = 0     , y - x , lemma₁ x y p
  ... | inr (inl p) = x - y , 0     , lemma₂ x y p
  ... | inr (inr p) = 0     , 0     , lemma₃ x y p

It remains to show that the procedure work respects the quotient. This is a truly gargantuan amount of work, and so it’s omitted from this page. You can unfold it below if you dare:

No, really, it’s quite ugly.
  -- I commend your bravery in unfolding this <details>! These three
  -- lemmas are inductively defined on the natural numbers in a way that
  -- lets us prove that the paths they return respect the Int quotient
  -- without using that Int is a set (because we don't know that yet!)

  lemma₁ zero    (suc y) p = refl
  lemma₁ (suc x) (suc y) p = lemma₁ x y (≤-peel p) ∙ Int.quot x y

  lemma₂ (suc x) zero    p = refl
  lemma₂ (suc x) (suc y) p = lemma₂ x y (≤-peel p) ∙ Int.quot x y

  lemma₃ zero zero p       = refl
  lemma₃ zero (suc y) p    = absurd (zero≠suc p)
  lemma₃ (suc x) zero p    = absurd (suc≠zero p)
  lemma₃ (suc x) (suc y) p = lemma₃ x y (suc-inj p) ∙ Int.quot x y

  abstract
    work-respects-quot
      : ∀ x y → PathP (λ i → Canonical (Int.quot x y i))
        (work x y) (work (suc x) (suc y))
    -- We split on (x, y) but also (1+x,1+y). This is obviously
    -- redundant to a human, but to Agda, we must do this: there is no
    -- link between these two splits.

    work-respects-quot x y with ≤-split x y | ≤-split (suc x) (suc y)
    ... | inl x<y | inl (s≤s x<y') = Σ-pathp refl $ Σ-pathp refl $
      commutes→square (∙-idl _)
    ... | inr (inl x>y) | inr (inl (s≤s x>y')) = Σ-pathp refl $ Σ-pathp refl $
      commutes→square (∙-idl _)
    ... | inr (inr x≡y) | inr (inr x≡y') = Σ-pathp refl $ Σ-pathp refl $
      commutes→square (∙-idl _)

    -- This *barrage* of cases is to handle the cases where e.g. (x < y)
    -- but (1 + x > 1 + y), which is "obviously" impossible. But Agda
    -- doesn't care about what humans think is obvious.
    ... | inl x<y | inr (inl (s≤s x>y))       = absurd (<-asym x<y x>y)
    ... | inl x<y | inr (inr x≡y)             = absurd (<-not-equal x<y (suc-inj x≡y))
    ... | inr (inl x>y) | inl (s≤s x<y)       = absurd (<-asym x>y x<y)
    ... | inr (inr x≡y) | inl (s≤s x<y)       = absurd (<-not-equal x<y x≡y)
    ... | inr (inl x>y) | inr (inr x≡y)       = absurd (<-not-equal x>y (sym (suc-inj x≡y)))
    ... | inr (inr x≡y) | inr (inl (s≤s x>y)) = absurd (<-irrefl (sym x≡y) x>y)

  go : ∀ n → Canonical n
  go (diff x y) = work x y
  go (Int.quot x y i) = work-respects-quot x y i

This immediately implies that the type of integers is a set, because it’s a retract of a set — namely

instance abstract
  H-Level-Int : ∀ {n} → H-Level Int (2 + n)
  H-Level-Int = basic-instance 2 (retract→is-hlevel 2 into from linv (hlevel 2)) where
    into : (Nat × Nat) → Int
    into (x , y) = diff x y

    from : Int → Nat × Nat
    from x with canonicalise x
    ... | a , b , p = a , b

    linv : ∀ x → into (from x) ≡ x
    linv x with canonicalise x
    ... | a , b , p = p

Recursion🔗

If we want to define a map it suffices to give a function which respects the quotient, in the following sense:

Int-rec : ∀ {ℓ} {X : Type ℓ}
        → (f : Nat → Nat → X)
        → (q : (a b : _) → f a b ≡ f (suc a) (suc b))
        → Int → X
Int-rec f q (diff x y) = f x y
Int-rec f q (quot m n i) = q m n i

However, since can be a more general space, not necessarily a set, defining a binary operation can be quite involved! It doesn’t suffice to exhibit a function from which respects the quotient separately in each argument:

Int-rec₂ : ∀ {ℓ} {B : Type ℓ}
         → (f : Nat × Nat → Nat × Nat → B)
         → (pl     : (a b x y : _) → f (a , b) (x , y) ≡ f (suc a , suc b) (x , y))
         → (pr     : (a b x y : _) → f (a , b) (x , y) ≡ f (a , b) (suc x , suc y))

In addition, we must have that these two paths pl and pr are coherent. There are two ways of obtaining an equality (pl after pr and pr after pl, respectively) and these must be homotopic:

         → (square : (a b x y : _) →
              Square (pl a b x y) (pr a b x y)
                     (pr (suc a) (suc b) x y)
                     (pl a b (suc x) (suc y)))
         → Int → Int → B

The type of square says that we need the following square of paths to commute, which says exactly that pl ∙ pr and pr ∙ pl are homotopic and imposes no further structure on 1:

Int-rec₂ f p-l p-r sq (diff a b) (diff x y)     = f (a , b) (x , y)
Int-rec₂ f p-l p-r sq (diff a b) (quot x y i)   = p-r a b x y i
Int-rec₂ f p-l p-r sq (quot a b i) (diff x y)   = p-l a b x y i
Int-rec₂ f p-l p-r sq (quot a b i) (quot x y j) = sq a b x y i j

However, when the type we are mapping into is a set, as is the case for the integers themselves, the square is automatically satisfied, so we can give a simplified recursion principle:

Int-rec₂-set :
  ∀ {ℓ} {B : Type ℓ} ⦃ iss-b : H-Level B 2 ⦄
  → (f : Nat × Nat → Nat × Nat → B)
  → (pl     : (a b x y : _) → f (a , b) (x , y) ≡ f (suc a , suc b) (x , y))
  → (pr     : (a b x y : _) → f (a , b) (x , y) ≡ f (a , b) (suc x , suc y))
  → Int → Int → B
Int-rec₂-set ⦃ iss-b ⦄ f pl pr = Int-rec₂ f pl pr square where abstract
  square
    : (a b x y : Nat)
    → PathP (λ i → pl a b x y i ≡ pl a b (suc x) (suc y) i)
            (pr a b x y) (pr (suc a) (suc b) x y)
  square a b x y = is-set→squarep (λ i j → hlevel 2) _ _ _ _

Furthermore, when proving propositions of the integers, the quotient is automatically respected, so it suffices to give the case for diff:

Int-elim-prop : ∀ {ℓ} {P : Int → Type ℓ}
              → ((x : Int) → is-prop (P x))
              → (f : (a b : Nat) → P (diff a b))
              → (x : Int) → P x
Int-elim-prop pprop f (diff a b) = f a b
Int-elim-prop pprop f (quot m n i) =
  is-prop→pathp (λ i → pprop (quot m n i)) (f m n) (f (suc m) (suc n)) i
There are also variants for binary and ternary predicates.
Int-elim₂-prop : ∀ {ℓ} {P : Int → Int → Type ℓ}
               → ((x y : Int) → is-prop (P x y))
               → (f : (a b x y : Nat) → P (diff a b) (diff x y))
               → (x : Int) (y : Int) → P x y
Int-elim₂-prop pprop f =
  Int-elim-prop (λ x → Π-is-hlevel 1 (pprop x))
    λ a b int → Int-elim-prop (λ x → pprop (diff a b) x) (f a b) int

Int-elim₃-prop : ∀ {ℓ} {P : Int → Int → Int → Type ℓ}
               → ((x y z : Int) → is-prop (P x y z))
               → (f : (a b c d e f : Nat) → P (diff a b) (diff c d) (diff e f))
               → (x : Int) (y : Int) (z : Int) → P x y z
Int-elim₃-prop pprop f =
  Int-elim₂-prop (λ x y → Π-is-hlevel 1 (pprop x y))
    λ a b c d int → Int-elim-prop (λ x → pprop (diff a b) (diff c d) x)
                                  (f a b c d)
                                  int

A third possibility for elimination is granted to us by the canonicalisation procedure: By canonicalising and looking at the resulting minuend and subtrahend2, we can split on an integer based on its sign:

Int-elim-by-sign
  : ∀ {ℓ} (P : Int → Type ℓ)
  → (pos : ∀ x → P (diff x 0))
  → (neg : ∀ x → P (diff 0 x))
  → P (diff 0 0)
  → ∀ x → P x
Int-elim-by-sign P pos neg zer x with inspect (canonicalise x)
... | (zero  , zero  , p) , q = subst P p zer
... | (zero  , suc b , p) , q = subst P p (neg (suc b))
... | (suc a , zero  , p) , q = subst P p (pos (suc a))
... | (suc a , suc b , p) , q =
  absurd (canonicalise-not-both-suc x (ap fst q) (ap (fst ∘ snd) q))

This procedure is useful, for instance, for computing absolute values:

abs : Int → Nat
abs x with by-sign x
... | posv z = z
... | negv z = suc z

_ : abs -10 ≡ 10
_ = refl

Algebra🔗

With these recursion and elimination helpers, it becomes routine to lift the algebraic operations from naturals to integers:

Successors🔗

The simplest “algebraic operation” on an integer is taking its successor. In fact, the integers are characterised by being the free type with an equivalence - that equivalence being “successor”.

sucℤ : Int → Int
sucℤ (diff x y)   = diff (suc x) y
sucℤ (quot m n i) = quot (suc m) n i

predℤ : Int → Int
predℤ (diff x y)   = diff x (suc y)
predℤ (quot m n i) = quot m (suc n) i

The successor of is Similarly, the predecessor of is By the generating equality quot, we have that predecessor and successor are inverses, since applying both (in either order) takes to

pred-sucℤ : (x : Int) → predℤ (sucℤ x) ≡ x
pred-sucℤ (diff x y)     = sym (quot x y)
pred-sucℤ (quot m n i) j = quot-diamond m n i (~ j)

suc-predℤ : (x : Int) → sucℤ (predℤ x) ≡ x
suc-predℤ (diff x y)     = sym (quot x y)
suc-predℤ (quot m n i) j = quot-diamond m n i (~ j)

sucℤ-is-equiv : is-equiv sucℤ
sucℤ-is-equiv = is-iso→is-equiv (iso predℤ suc-predℤ pred-sucℤ)

predℤ-is-equiv : is-equiv predℤ
predℤ-is-equiv = is-iso→is-equiv (iso sucℤ pred-sucℤ suc-predℤ)

Addition🔗

_+ℤ_ : Int → Int → Int
diff a b +ℤ diff c d = diff (a + c) (b + d)
diff x y +ℤ quot m n i =
  (quot (x + m) (y + n) ∙ sym (ap₂ diff (+-sucr x m) (+-sucr y n))) i
quot m n i +ℤ diff x y = quot (m + x) (n + y) i
quot m n i +ℤ quot m' n' j =
  is-set→squarep (λ i j → hlevel 2)
    (λ j → quot (m + m') (n + n') j)
    (quot (m + m') (n + n') ∙ sym (ap₂ diff (+-sucr m m') (+-sucr n n')))
    ( quot (suc (m + m')) (suc (n + n'))
    ∙ sym (ap₂ diff (ap suc (+-sucr m m')) (ap suc (+-sucr n n'))))
    (λ j → quot (m + suc m') (n + suc n') j)
    i j

Since addition of integers is (essentially!) addition of pairs of naturals, the algebraic properties of + on the natural numbers automatically lift to properties about _+ℤ_, using the recursion helpers for props (Int-elim-prop) and the fact that equality of integers is a proposition.

+ℤ-associative : (x y z : Int) → x +ℤ (y +ℤ z) ≡ (x +ℤ y) +ℤ z
+ℤ-zerol       : (x : Int)     → 0 +ℤ x ≡ x
+ℤ-zeror       : (x : Int)     → x +ℤ 0 ≡ x
+ℤ-commutative : (x y : Int)   → x +ℤ y ≡ y +ℤ x
See the proofs here
abstract
  +ℤ-associative =
    Int-elim₃-prop
      (λ x y z → hlevel 1)
      (λ a b c d e f → ap₂ diff (+-associative a c e) (+-associative b d f))
  +ℤ-zerol = Int-elim-prop (λ x → hlevel 1) (λ a b → refl)
  +ℤ-zeror =
    Int-elim-prop (λ x → hlevel 1) (λ a b → ap₂ diff (+-zeror a) (+-zeror b))
  +ℤ-commutative =
    Int-elim₂-prop (λ x y → hlevel 1)
      (λ a b c d → ap₂ diff (+-commutative a c) (+-commutative b d))

Inverses🔗

Every integer has an additive inverse, denoted which is obtained by swapping the components of the pair. Since the definition of negate is very simple, it can be written conveniently without using Int-rec:

negate : Int → Int
negate (diff x y) = diff y x
negate (quot m n i) = quot n m i

The proof that is an additive inverse to follows, essentially, from commutativity of addition on natural numbers, and the fact that all zeroes are identified.

abstract
  +ℤ-inverser : (x : Int) → x +ℤ negate x ≡ 0
  +ℤ-inverser =
    Int-elim-prop (λ _ → hlevel 1) λ where
      a b → diff (a + b) ⌜ b + a ⌝ ≡⟨ ap! (+-commutative b a) ⟩≡
            diff (a + b) (a + b)   ≡⟨ sym (zeroes (a + b)) ⟩≡
            diff 0 0               ∎

  +ℤ-inversel : (x : Int) → negate x +ℤ x ≡ 0
  +ℤ-inversel =
    Int-elim-prop (λ _ → hlevel 1) λ where
      a b → diff ⌜ b + a ⌝ (a + b) ≡⟨ ap! (+-commutative b a) ⟩≡
            diff (a + b) (a + b)   ≡⟨ sym (zeroes (a + b)) ⟩≡
            diff 0 0               ∎

Since negate is precisely what’s missing for Nat to be a group, we can turn the integers into a group. Subtraction is defined as addition with the inverse, rather than directly on diff:

_-ℤ_ : Int → Int → Int
x -ℤ y = x +ℤ negate y

Multiplication🔗

We now prove that the integers are a ring, i.e. that there is a multiplication operation with 1 as a left/right identity, which is associative, and additionally distributes over addition on both the left and the right. It’s also commutative — so is a commutative ring.

The definition of multiplication is slightly tricky: We use the binomial theorem. Pretend that is really and expand that to that’s our product. It remains to show that this respects the defining equation quot, which involves some nasty equations (you can see them in the types of l₁ and l₂ below) — but this can be done with the semiring solver.

  _*ℤ_ : Int → Int → Int
  diff a b *ℤ diff c d = diff (a * c + b * d) (a * d + b * c)
  diff x y *ℤ quot m n i = rhs₁ x y m n i
  quot m n i *ℤ diff x y = rhs₂ x y m n i
  quot m n i *ℤ quot x y j = is-set→squarep (λ i j → hlevel 2)
    (rhs₂ x y m n) (rhs₁ m n x y)
    (rhs₁ (suc m) (suc n) x y) (rhs₂ (suc x) (suc y) m n) i j

We omit the proofs of the arithmetic identities below since they are essentially induction + calling the semiring solver.

abstract
  *ℤ-associative : ∀ x y z → x *ℤ (y *ℤ z) ≡ (x *ℤ y) *ℤ z
  *ℤ-commutative : ∀ x y → x *ℤ y ≡ y *ℤ x
  *ℤ-idl : ∀ x → 1 *ℤ x ≡ x
  *ℤ-idr : ∀ x → x *ℤ 1 ≡ x
  *ℤ-distrib-+ℤ-l : ∀ x y z → x *ℤ (y +ℤ z) ≡ (x *ℤ y) +ℤ (x *ℤ z)
  *ℤ-distrib-+ℤ-r : ∀ x y z → (y +ℤ z) *ℤ x ≡ (y *ℤ x) +ℤ (z *ℤ x)
  *ℤ-associative =
    Int-elim₃-prop (λ _ _ _ → hlevel 1)
      λ a b c d e f → sym (same-difference (lemma a b c d e f))
    where abstract
      lemma
        : ∀ a b c d e f
        → (a * c + b * d) * e + (a * d + b * c) * f + (a * (c * f + d * e) + b * (c * e + d * f))
        ≡ (a * c + b * d) * f + (a * d + b * c) * e + (a * (c * e + d * f) + b * (c * f + d * e))
      lemma a b c d e f = nat!

  *ℤ-commutative =
    Int-elim₂-prop (λ _ _ → hlevel 1) λ a b x y → same-difference (lemma a b x y)
    where abstract
      lemma : ∀ a b x y → a * x + b * y + (x * b + y * a) ≡ a * y + b * x + (x * a + y * b)
      lemma a b x y = nat!

  *ℤ-idl = Int-elim-prop (λ _ → hlevel 1) (λ a b → same-difference (lemma a b))
    where abstract
      lemma : ∀ a b → (a + 0 + 0 + b) ≡ (b + 0 + 0 + a)
      lemma a b = nat!

  *ℤ-idr = Int-elim-prop (λ _ → hlevel 1) (λ a b → same-difference (lemma a b))
    where abstract
      lemma : ∀ a b → a * 1 + b * 0 + b ≡ a * 0 + b * 1 + a
      lemma a b = nat!

  *ℤ-distrib-+ℤ-l = Int-elim₃-prop (λ _ _ _ → hlevel 1)
    λ a b c d e f → same-difference (lemma a b c d e f)
    where abstract
      lemma : ∀ a b c d e f → a * (c + e) + b * (d + f) + (a * d + b * c + (a * f + b * e)) ≡ a * (d + f) + b * (c + e) + (a * c + b * d + (a * e + b * f))
      lemma a b c d e f = nat!

  *ℤ-distrib-+ℤ-r = Int-elim₃-prop (λ _ _ _ → hlevel 1)
    λ a b c d e f → same-difference (lemma a b c d e f)
    where
      lemma : ∀ a b c d e f → (c + e) * a + (d + f) * b + (c * b + d * a + (e * b + f * a)) ≡ (c + e) * b + (d + f) * a + (c * a + d * b + (e * a + f * b))
      lemma a b c d e f = nat!

canonicalise-injective
  : ∀ x y
  → canonicalise x .fst      ≡ canonicalise y .fst
  → canonicalise x .snd .fst ≡ canonicalise y .snd .fst
  → x ≡ y
canonicalise-injective = Int-elim₂-prop (λ _ _ → hlevel 1) λ a b x y p q →
     sym (canonicalise (diff a b) .snd .snd)
  ·· ap₂ diff p q
  ·· canonicalise (diff x y) .snd .snd

instance
  Discrete-Int : Discrete Int
  Discrete-Int = go _ _ where
    go₀ : (a b x y : Nat) → Dec (diff a b ≡ diff x y)
    go₀ a b x y with inspect (a + y == b + x)
    ... | true , p  = yes (same-difference (is-equal→path {a + y} {b + x} p))
    ... | false , p = no λ q → is-not-equal→not-path
      {a + y} {b + x} p (ℤ-Path.encode a b (diff x y) q)

    go₁ : (a b : Nat) (y : Int) → Dec (diff a b ≡ y)
    go₁ a b (diff x y) = go₀ a b x y
    go₁ a b (quot m n i) = is-prop→pathp
      (λ i → Dec-is-hlevel 1 (hlevel {T = diff a b ≡ quot m n i} 1))
      (go₀ a b m n)
      (go₀ a b (suc m) (suc n)) i

    go : ∀ x y → Dec (x ≡ y)
    go (diff a b) y = go₁ a b y
    go (quot m n i) y = is-prop→pathp
      (λ i → Dec-is-hlevel 1 (hlevel {T = quot m n i ≡ y} 1))
      (go₁ m n y)
      (go₁ (suc m) (suc n) y) i

  1. In the diagram, we write for suc x.↩︎

  2. keeping in mind that the image of canonicalise is restricted to pairs of the form or ↩︎