Metamath Proof Explorer


Theorem sdom1

Description: A set has less than one member iff it is empty. (Contributed by Stefan O'Rear, 28-Oct-2014) Avoid ax-pow , ax-un . (Revised by BTernaryTau, 12-Dec-2024)

Ref Expression
Assertion sdom1 ⊢ A ≺ 1 𝑜 ↔ A = ∅

Proof

Step Hyp Ref Expression
1 df1o2 ⊢ 1 𝑜 = ∅
2 1 breq2i ⊢ A ≼ 1 𝑜 ↔ A ≼ ∅
3 brdomi ⊢ A ≼ ∅ → ∃ f f : A ⟶ 1-1 ∅
4 f1cdmsn ⊢ f : A ⟶ 1-1 ∅ ∧ A ≠ ∅ → ∃ x A = x
5 vex ⊢ x ∈ V
6 5 ensn1 ⊢ x ≈ 1 𝑜
7 breq1 ⊢ A = x → A ≈ 1 𝑜 ↔ x ≈ 1 𝑜
8 6 7 mpbiri ⊢ A = x → A ≈ 1 𝑜
9 8 exlimiv ⊢ ∃ x A = x → A ≈ 1 𝑜
10 4 9 syl ⊢ f : A ⟶ 1-1 ∅ ∧ A ≠ ∅ → A ≈ 1 𝑜
11 10 expcom ⊢ A ≠ ∅ → f : A ⟶ 1-1 ∅ → A ≈ 1 𝑜
12 11 exlimdv ⊢ A ≠ ∅ → ∃ f f : A ⟶ 1-1 ∅ → A ≈ 1 𝑜
13 3 12 syl5 ⊢ A ≠ ∅ → A ≼ ∅ → A ≈ 1 𝑜
14 2 13 biimtrid ⊢ A ≠ ∅ → A ≼ 1 𝑜 → A ≈ 1 𝑜
15 iman ⊢ A ≼ 1 𝑜 → A ≈ 1 𝑜 ↔ ¬ A ≼ 1 𝑜 ∧ ¬ A ≈ 1 𝑜
16 14 15 sylib ⊢ A ≠ ∅ → ¬ A ≼ 1 𝑜 ∧ ¬ A ≈ 1 𝑜
17 brsdom ⊢ A ≺ 1 𝑜 ↔ A ≼ 1 𝑜 ∧ ¬ A ≈ 1 𝑜
18 16 17 sylnibr ⊢ A ≠ ∅ → ¬ A ≺ 1 𝑜
19 18 necon4ai ⊢ A ≺ 1 𝑜 → A = ∅
20 1n0 ⊢ 1 𝑜 ≠ ∅
21 1oex ⊢ 1 𝑜 ∈ V
22 21 0sdom ⊢ ∅ ≺ 1 𝑜 ↔ 1 𝑜 ≠ ∅
23 20 22 mpbir ⊢ ∅ ≺ 1 𝑜
24 breq1 ⊢ A = ∅ → A ≺ 1 𝑜 ↔ ∅ ≺ 1 𝑜
25 23 24 mpbiri ⊢ A = ∅ → A ≺ 1 𝑜
26 19 25 impbii ⊢ A ≺ 1 𝑜 ↔ A = ∅