Metamath Proof Explorer


Theorem hashfz1

Description: The set ( 1 ... N ) has N elements. (Contributed by Paul Chapman, 22-Jun-2011) (Revised by Mario Carneiro, 15-Sep-2013)

Ref Expression
Assertion hashfz1 ⊢ N ∈ ℕ 0 → 1 … N = N

Proof

Step Hyp Ref Expression
1 eqid ⊢ rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω = rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω
2 1 cardfz ⊢ N ∈ ℕ 0 → card ⁡ 1 … N = rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω -1 ⁡ N
3 2 fveq2d ⊢ N ∈ ℕ 0 → rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω ⁡ card ⁡ 1 … N = rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω ⁡ rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω -1 ⁡ N
4 fzfid ⊢ N ∈ ℕ 0 → 1 … N ∈ Fin
5 1 hashgval ⊢ 1 … N ∈ Fin → rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω ⁡ card ⁡ 1 … N = 1 … N
6 4 5 syl ⊢ N ∈ ℕ 0 → rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω ⁡ card ⁡ 1 … N = 1 … N
7 1 hashgf1o ⊢ rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω : ω ⟶ 1-1 onto ℕ 0
8 f1ocnvfv2 ⊢ rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω : ω ⟶ 1-1 onto ℕ 0 ∧ N ∈ ℕ 0 → rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω ⁡ rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω -1 ⁡ N = N
9 7 8 mpan ⊢ N ∈ ℕ 0 → rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω ⁡ rec ⁡ x ∈ V ⟼ x + 1 0 ↾ ω -1 ⁡ N = N
10 3 6 9 3eqtr3d ⊢ N ∈ ℕ 0 → 1 … N = N