Metamath Proof Explorer


Theorem hashomiso

Description: The # function yields an order isomorphism between _om and NN0 . (Contributed by Eric Schmidt, 7-Jul-2026)

Ref Expression
Assertion hashomiso ⊢ . ↾ ω Isom E , < ω ℕ 0

Proof

Step Hyp Ref Expression
1 hashomf1o ⊢ . ↾ ω : ω ⟶ 1-1 onto ℕ 0
2 epel ⊢ x E y ↔ x ∈ y
3 hashnnltb ⊢ x ∈ ω ∧ y ∈ ω → x ∈ y ↔ x < y
4 2 3 bitrid ⊢ x ∈ ω ∧ y ∈ ω → x E y ↔ x < y
5 fvres ⊢ x ∈ ω → . ↾ ω ⁡ x = x
6 fvres ⊢ y ∈ ω → . ↾ ω ⁡ y = y
7 5 6 breqan12d ⊢ x ∈ ω ∧ y ∈ ω → . ↾ ω ⁡ x < . ↾ ω ⁡ y ↔ x < y
8 4 7 bitr4d ⊢ x ∈ ω ∧ y ∈ ω → x E y ↔ . ↾ ω ⁡ x < . ↾ ω ⁡ y
9 8 rgen2 ⊢ ∀ x ∈ ω ∀ y ∈ ω x E y ↔ . ↾ ω ⁡ x < . ↾ ω ⁡ y
10 df-isom ⊢ . ↾ ω Isom E , < ω ℕ 0 ↔ . ↾ ω : ω ⟶ 1-1 onto ℕ 0 ∧ ∀ x ∈ ω ∀ y ∈ ω x E y ↔ . ↾ ω ⁡ x < . ↾ ω ⁡ y
11 1 9 10 mpbir2an ⊢ . ↾ ω Isom E , < ω ℕ 0