Chuangjie Xu 2011, with changes by Martin Escardo later.
\begin{code}
{-# OPTIONS --safe --without-K #-}
module Naturals.Addition where
open import MLTT.Spartan hiding (_+_)
open import Naturals.Properties
infixl 31 _+_
_+_ : ℕ → ℕ → ℕ
n + 0 = n
n + (succ m) = succ (n + m)
_+ᴸ_ : ℕ → ℕ → ℕ
m +ᴸ n = n + m
{-# BUILTIN NATPLUS _+_ #-}
zero-right-neutral : (n : ℕ) → n + 0 ＝ n
zero-right-neutral n = refl
zero-left-neutral : (n : ℕ) → 0 + n ＝ n
zero-left-neutral = ℕ-induction base step
where
base : 0 + 0 ＝ 0
base = refl
step : (n : ℕ) → 0 + n ＝ n → 0 + succ n ＝ succ n
step n IH = 0 + succ n ＝⟨ refl ⟩
succ (0 + n) ＝⟨ ap succ IH ⟩
succ n ∎
addition-associativity : (l n m : ℕ) → (l + n) + m ＝ l + (n + m)
addition-associativity l n = ℕ-induction base step
where
base : (l + n) + 0 ＝ l + (n + 0)
base = (l + n) + 0 ＝⟨ refl ⟩
l + n ＝⟨ refl ⟩
l + (n + 0) ∎
step : (m : ℕ) → (l + n) + m ＝ l + (n + m)
→ (l + n) + succ m ＝ l + (n + succ m)
step m IH = (l + n) + succ m ＝⟨ refl ⟩
succ ((l + n) + m) ＝⟨ ap succ IH ⟩
succ (l + (n + m)) ＝⟨ refl ⟩
l + succ (n + m) ＝⟨ refl ⟩
l + (n + succ m) ∎
addition-commutativity : (n m : ℕ) → n + m ＝ m + n
addition-commutativity n = ℕ-induction base step
where
base : n + 0 ＝ 0 + n
base = n + 0 ＝⟨ zero-right-neutral n ⟩
n ＝⟨ (zero-left-neutral n)⁻¹ ⟩
0 + n ∎
step : (m : ℕ) → n + m ＝ m + n → n + succ m ＝ succ m + n
step m IH = n + succ m ＝⟨ refl ⟩
succ (n + m) ＝⟨ ap succ IH ⟩
succ (m + n) ＝⟨ lemma₀ (m + n) ⟩
1 + (m + n) ＝⟨ (addition-associativity 1 m n)⁻¹ ⟩
(1 + m) + n ＝⟨ ap (_+ n) ((lemma₀ m)⁻¹) ⟩
succ m + n ∎
where
lemma₀ : (k : ℕ) → succ k ＝ 1 + k
lemma₀ = ℕ-induction base₀ step₀
where
base₀ : succ 0 ＝ 1 + 0
base₀ = refl
step₀ : (k : ℕ) → succ k ＝ 1 + k → succ (succ k) ＝ 1 + succ k
step₀ k IH = succ (succ k) ＝⟨ ap succ IH ⟩
succ (1 + k) ＝⟨ refl ⟩
1 + succ k ∎
trivial-addition-rearrangement : (x y z : ℕ) → x + y + z ＝ x + z + y
trivial-addition-rearrangement x y z =
(x + y) + z ＝⟨ addition-associativity x y z ⟩
x + (y + z) ＝⟨ ap (x +_) (addition-commutativity y z) ⟩
x + (z + y) ＝⟨ (addition-associativity x z y)⁻¹ ⟩
(x + z) + y ∎
\end{code}
Added 12/05/2020 by Andrew Sneap.
Some more simple properties of addition. The proofs all use induction,
apart from addition-right-cancellable which follows from
addition-left-cancellable and commutativity of addition.
\begin{code}
succ-right : (x y : ℕ) → x + succ y ＝ succ (x + y)
succ-right x y = refl
succ-left : (x y : ℕ) → succ x + y ＝ succ (x + y)
succ-left x = ℕ-induction base step
where
base : succ x + 0 ＝ succ (x + 0)
base = succ x + 0 ＝⟨ refl ⟩
succ x ＝⟨ ap succ refl ⟩
succ (x + 0) ∎
step : (k : ℕ)
→ succ x + k ＝ succ (x + k)
→ succ x + succ k ＝ succ (x + succ k)
step k IH = succ x + succ k ＝⟨ refl ⟩
succ (succ x + k) ＝⟨ ap succ IH ⟩
succ (succ (x + k)) ＝⟨ refl ⟩
succ (x + succ k) ∎
addition-left-cancellable : (x y z : ℕ) → z + x ＝ z + y → x ＝ y
addition-left-cancellable x y = ℕ-induction base step
where
base : 0 + x ＝ 0 + y → x ＝ y
base h = x ＝⟨ zero-left-neutral x ⁻¹ ⟩
0 + x ＝⟨ h ⟩
0 + y ＝⟨ zero-left-neutral y ⟩
y ∎
step : (k : ℕ)
→ (k + x ＝ k + y → x ＝ y)
→ (succ k + x ＝ succ k + y → x ＝ y)
step k IH r = IH (succ-lc (lemma₁ r))
where
lemma₁ : succ k + x ＝ succ k + y → succ (k + x) ＝ succ (k + y)
lemma₁ r = succ (k + x) ＝⟨ succ-left k x ⁻¹ ⟩
succ k + x ＝⟨ r ⟩
succ k + y ＝⟨ succ-left k y ⟩
succ (k + y) ∎
addition-right-cancellable : (x y z : ℕ) → x + z ＝ y + z → x ＝ y
addition-right-cancellable x y z r = addition-left-cancellable x y z lemma₀
where
lemma₀ : z + x ＝ z + y
lemma₀ = z + x ＝⟨ addition-commutativity z x ⟩
x + z ＝⟨ r ⟩
y + z ＝⟨ addition-commutativity y z ⟩
z + y ∎
\end{code}
We also have that if the sum of two numbers are zero, then the right
argument is zero. The left argument is zero, which can be proved using
commutativity of addition. This function is needed in the HCF file.
\begin{code}
sum-to-zero-gives-zero : (x y : ℕ) → x + y ＝ 0 → y ＝ 0
sum-to-zero-gives-zero x 0 e = refl
sum-to-zero-gives-zero x (succ y) e = 𝟘-elim (positive-not-zero (x + y) e)
\end{code}