Metamath Proof Explorer


Theorem 0sdom1dom

Description: Strict dominance over 0 is the same as dominance over 1. For a shorter proof requiring ax-un , see 0sdom1domALT . (Contributed by NM, 28-Sep-2004) Avoid ax-un . (Revised by BTernaryTau, 7-Dec-2024)

Ref Expression
Assertion 0sdom1dom ⊢ ∅ ≺ A ↔ 1 𝑜 ≼ A

Proof

Step Hyp Ref Expression
1 relsdom ⊢ Rel ⁡ ≺
2 1 brrelex2i ⊢ ∅ ≺ A → A ∈ V
3 reldom ⊢ Rel ⁡ ≼
4 3 brrelex2i ⊢ 1 𝑜 ≼ A → A ∈ V
5 0sdomg ⊢ A ∈ V → ∅ ≺ A ↔ A ≠ ∅
6 n0 ⊢ A ≠ ∅ ↔ ∃ x x ∈ A
7 snssi ⊢ x ∈ A → x ⊆ A
8 df1o2 ⊢ 1 𝑜 = ∅
9 0ex ⊢ ∅ ∈ V
10 vex ⊢ x ∈ V
11 en2sn ⊢ ∅ ∈ V ∧ x ∈ V → ∅ ≈ x
12 9 10 11 mp2an ⊢ ∅ ≈ x
13 8 12 eqbrtri ⊢ 1 𝑜 ≈ x
14 endom ⊢ 1 𝑜 ≈ x → 1 𝑜 ≼ x
15 13 14 ax-mp ⊢ 1 𝑜 ≼ x
16 domssr ⊢ A ∈ V ∧ x ⊆ A ∧ 1 𝑜 ≼ x → 1 𝑜 ≼ A
17 15 16 mp3an3 ⊢ A ∈ V ∧ x ⊆ A → 1 𝑜 ≼ A
18 17 ex ⊢ A ∈ V → x ⊆ A → 1 𝑜 ≼ A
19 7 18 syl5 ⊢ A ∈ V → x ∈ A → 1 𝑜 ≼ A
20 19 exlimdv ⊢ A ∈ V → ∃ x x ∈ A → 1 𝑜 ≼ A
21 6 20 biimtrid ⊢ A ∈ V → A ≠ ∅ → 1 𝑜 ≼ A
22 1n0 ⊢ 1 𝑜 ≠ ∅
23 dom0 ⊢ 1 𝑜 ≼ ∅ ↔ 1 𝑜 = ∅
24 22 23 nemtbir ⊢ ¬ 1 𝑜 ≼ ∅
25 breq2 ⊢ A = ∅ → 1 𝑜 ≼ A ↔ 1 𝑜 ≼ ∅
26 24 25 mtbiri ⊢ A = ∅ → ¬ 1 𝑜 ≼ A
27 26 necon2ai ⊢ 1 𝑜 ≼ A → A ≠ ∅
28 21 27 impbid1 ⊢ A ∈ V → A ≠ ∅ ↔ 1 𝑜 ≼ A
29 5 28 bitrd ⊢ A ∈ V → ∅ ≺ A ↔ 1 𝑜 ≼ A
30 2 4 29 pm5.21nii ⊢ ∅ ≺ A ↔ 1 𝑜 ≼ A