```
{-# OPTIONS --cubical #-}

----------

A = ℕ

P : A → Type
P n = Vec n

_+_ 
zero + y = y
suc x + y = suc (x + y)

-- x and y are NOT definitionally equal
x = 1 + 2
y = 2 + 1

-- 
p : x ≡ y
p = +-comm 1 2


Identity {A : Type} {x y : A}

elim-unit :
  (A : Type) →

  -- unit case
  (a : Unit → A) →

  Unit → A



elim-nat : (A : Type) → (z : A) → (s : A → A) → ℕ → A


elim-nat A f g 5 = a5
  - a0 = f
  - a1 = g a0
  - a2 = g a1
  - a3 = g a2
  - a4 = g a3
  - a5 = g a4



elim-identity :
  -- type that the identity type is over
  (A : Type) → 

  -- output type
  (C : (x y : A) → (p : x ≡ y) → Type) →

  -- refl case
  (c : (z : A) → C z z (refl z)) →

  (a b : A) → (p′ : a ≡ b) →
  C a b p′



-- In the case of transports
D is C in the eliminator

P : A → Type

D : (x y : A) → (p : x ≡ y) → Type
D x y p = P x → P y

transport : (A : Type) → (P : A → Type) → (x y : A) → (p : x ≡ y) → P x → P y
transport = elim-identity A (λ a b _ → (P a → P b)) (λ a → id) x y p


elim-identity A D d (a a : A) (refl a) = d a


transport A P x x (refl x)
  = elim-identity A (λ a b _ → (P a → P b)) (λ x → id) x x (refl x)
  = (λ x → id) x
  = id

---


-- Goal:
p* : P (1 + 2) ≡ P (2 + 1)
























----------

loop ∙ loop ∙ loop : S¹
3 : ℤ

p : {A : Type} {x y : A} → x ≡ y
-- Path from P base = ℤ to P base

P : S¹ → Type
P base = ℤ
P loop = ?
```

For this week:

- Try to understand Lemma 2.3.4 in HoTT book