Metamath Proof Explorer


Theorem en1

Description: A set is equinumerous to ordinal one iff it is a singleton. (Contributed by NM, 25-Jul-2004) Avoid ax-un . (Revised by BTernaryTau, 23-Sep-2024)

Ref Expression
Assertion en1 ⊢ A ≈ 1 𝑜 ↔ ∃ x A = x

Proof

Step Hyp Ref Expression
1 df1o2 ⊢ 1 𝑜 = ∅
2 1 breq2i ⊢ A ≈ 1 𝑜 ↔ A ≈ ∅
3 encv ⊢ A ≈ ∅ → A ∈ V ∧ ∅ ∈ V
4 breng ⊢ A ∈ V ∧ ∅ ∈ V → A ≈ ∅ ↔ ∃ f f : A ⟶ 1-1 onto ∅
5 3 4 syl ⊢ A ≈ ∅ → A ≈ ∅ ↔ ∃ f f : A ⟶ 1-1 onto ∅
6 5 ibi ⊢ A ≈ ∅ → ∃ f f : A ⟶ 1-1 onto ∅
7 2 6 sylbi ⊢ A ≈ 1 𝑜 → ∃ f f : A ⟶ 1-1 onto ∅
8 f1ocnv ⊢ f : A ⟶ 1-1 onto ∅ → f -1 : ∅ ⟶ 1-1 onto A
9 f1ofo ⊢ f -1 : ∅ ⟶ 1-1 onto A → f -1 : ∅ ⟶ onto A
10 forn ⊢ f -1 : ∅ ⟶ onto A → ran ⁡ f -1 = A
11 9 10 syl ⊢ f -1 : ∅ ⟶ 1-1 onto A → ran ⁡ f -1 = A
12 f1of ⊢ f -1 : ∅ ⟶ 1-1 onto A → f -1 : ∅ ⟶ A
13 0ex ⊢ ∅ ∈ V
14 13 fsn2 ⊢ f -1 : ∅ ⟶ A ↔ f -1 ⁡ ∅ ∈ A ∧ f -1 = ∅ f -1 ⁡ ∅
15 14 simprbi ⊢ f -1 : ∅ ⟶ A → f -1 = ∅ f -1 ⁡ ∅
16 12 15 syl ⊢ f -1 : ∅ ⟶ 1-1 onto A → f -1 = ∅ f -1 ⁡ ∅
17 16 rneqd ⊢ f -1 : ∅ ⟶ 1-1 onto A → ran ⁡ f -1 = ran ⁡ ∅ f -1 ⁡ ∅
18 13 rnsnop ⊢ ran ⁡ ∅ f -1 ⁡ ∅ = f -1 ⁡ ∅
19 17 18 eqtrdi ⊢ f -1 : ∅ ⟶ 1-1 onto A → ran ⁡ f -1 = f -1 ⁡ ∅
20 11 19 eqtr3d ⊢ f -1 : ∅ ⟶ 1-1 onto A → A = f -1 ⁡ ∅
21 fvex ⊢ f -1 ⁡ ∅ ∈ V
22 sneq ⊢ x = f -1 ⁡ ∅ → x = f -1 ⁡ ∅
23 22 eqeq2d ⊢ x = f -1 ⁡ ∅ → A = x ↔ A = f -1 ⁡ ∅
24 21 23 spcev ⊢ A = f -1 ⁡ ∅ → ∃ x A = x
25 8 20 24 3syl ⊢ f : A ⟶ 1-1 onto ∅ → ∃ x A = x
26 25 exlimiv ⊢ ∃ f f : A ⟶ 1-1 onto ∅ → ∃ x A = x
27 7 26 syl ⊢ A ≈ 1 𝑜 → ∃ x A = x
28 vex ⊢ x ∈ V
29 28 ensn1 ⊢ x ≈ 1 𝑜
30 breq1 ⊢ A = x → A ≈ 1 𝑜 ↔ x ≈ 1 𝑜
31 29 30 mpbiri ⊢ A = x → A ≈ 1 𝑜
32 31 exlimiv ⊢ ∃ x A = x → A ≈ 1 𝑜
33 27 32 impbii ⊢ A ≈ 1 𝑜 ↔ ∃ x A = x