Metamath Proof Explorer


Theorem hashen1

Description: A set has size 1 if and only if it is equinumerous to the ordinal 1. (Contributed by AV, 14-Apr-2019)

Ref Expression
Assertion hashen1 ⊢ A ∈ V → A = 1 ↔ A ≈ 1 𝑜

Proof

Step Hyp Ref Expression
1 0ex ⊢ ∅ ∈ V
2 hashsng ⊢ ∅ ∈ V → ∅ = 1
3 1 2 ax-mp ⊢ ∅ = 1
4 3 eqcomi ⊢ 1 = ∅
5 4 a1i ⊢ A ∈ V → 1 = ∅
6 5 eqeq2d ⊢ A ∈ V → A = 1 ↔ A = ∅
7 simpr ⊢ A ∈ V ∧ A = ∅ → A = ∅
8 1nn0 ⊢ 1 ∈ ℕ 0
9 3 8 eqeltri ⊢ ∅ ∈ ℕ 0
10 hashvnfin ⊢ A ∈ V ∧ ∅ ∈ ℕ 0 → A = ∅ → A ∈ Fin
11 9 10 mpan2 ⊢ A ∈ V → A = ∅ → A ∈ Fin
12 11 imp ⊢ A ∈ V ∧ A = ∅ → A ∈ Fin
13 snfi ⊢ ∅ ∈ Fin
14 hashen ⊢ A ∈ Fin ∧ ∅ ∈ Fin → A = ∅ ↔ A ≈ ∅
15 12 13 14 sylancl ⊢ A ∈ V ∧ A = ∅ → A = ∅ ↔ A ≈ ∅
16 7 15 mpbid ⊢ A ∈ V ∧ A = ∅ → A ≈ ∅
17 16 ex ⊢ A ∈ V → A = ∅ → A ≈ ∅
18 hasheni ⊢ A ≈ ∅ → A = ∅
19 17 18 impbid1 ⊢ A ∈ V → A = ∅ ↔ A ≈ ∅
20 df1o2 ⊢ 1 𝑜 = ∅
21 20 eqcomi ⊢ ∅ = 1 𝑜
22 21 breq2i ⊢ A ≈ ∅ ↔ A ≈ 1 𝑜
23 22 a1i ⊢ A ∈ V → A ≈ ∅ ↔ A ≈ 1 𝑜
24 6 19 23 3bitrd ⊢ A ∈ V → A = 1 ↔ A ≈ 1 𝑜