Metamath Proof Explorer


Theorem 2dom

Description: A set that dominates ordinal 2 has at least 2 different members. (Contributed by NM, 25-Jul-2004)

Ref Expression
Assertion 2dom ⊢ 2 𝑜 ≼ A → ∃ x ∈ A ∃ y ∈ A ¬ x = y

Proof

Step Hyp Ref Expression
1 df2o2 ⊢ 2 𝑜 = ∅ ∅
2 1 breq1i ⊢ 2 𝑜 ≼ A ↔ ∅ ∅ ≼ A
3 brdomi ⊢ ∅ ∅ ≼ A → ∃ f f : ∅ ∅ ⟶ 1-1 A
4 2 3 sylbi ⊢ 2 𝑜 ≼ A → ∃ f f : ∅ ∅ ⟶ 1-1 A
5 f1f ⊢ f : ∅ ∅ ⟶ 1-1 A → f : ∅ ∅ ⟶ A
6 0ex ⊢ ∅ ∈ V
7 6 prid1 ⊢ ∅ ∈ ∅ ∅
8 ffvelcdm ⊢ f : ∅ ∅ ⟶ A ∧ ∅ ∈ ∅ ∅ → f ⁡ ∅ ∈ A
9 5 7 8 sylancl ⊢ f : ∅ ∅ ⟶ 1-1 A → f ⁡ ∅ ∈ A
10 snex ⊢ ∅ ∈ V
11 10 prid2 ⊢ ∅ ∈ ∅ ∅
12 ffvelcdm ⊢ f : ∅ ∅ ⟶ A ∧ ∅ ∈ ∅ ∅ → f ⁡ ∅ ∈ A
13 5 11 12 sylancl ⊢ f : ∅ ∅ ⟶ 1-1 A → f ⁡ ∅ ∈ A
14 0nep0 ⊢ ∅ ≠ ∅
15 14 neii ⊢ ¬ ∅ = ∅
16 f1fveq ⊢ f : ∅ ∅ ⟶ 1-1 A ∧ ∅ ∈ ∅ ∅ ∧ ∅ ∈ ∅ ∅ → f ⁡ ∅ = f ⁡ ∅ ↔ ∅ = ∅
17 7 11 16 mpanr12 ⊢ f : ∅ ∅ ⟶ 1-1 A → f ⁡ ∅ = f ⁡ ∅ ↔ ∅ = ∅
18 15 17 mtbiri ⊢ f : ∅ ∅ ⟶ 1-1 A → ¬ f ⁡ ∅ = f ⁡ ∅
19 eqeq1 ⊢ x = f ⁡ ∅ → x = y ↔ f ⁡ ∅ = y
20 19 notbid ⊢ x = f ⁡ ∅ → ¬ x = y ↔ ¬ f ⁡ ∅ = y
21 eqeq2 ⊢ y = f ⁡ ∅ → f ⁡ ∅ = y ↔ f ⁡ ∅ = f ⁡ ∅
22 21 notbid ⊢ y = f ⁡ ∅ → ¬ f ⁡ ∅ = y ↔ ¬ f ⁡ ∅ = f ⁡ ∅
23 20 22 rspc2ev ⊢ f ⁡ ∅ ∈ A ∧ f ⁡ ∅ ∈ A ∧ ¬ f ⁡ ∅ = f ⁡ ∅ → ∃ x ∈ A ∃ y ∈ A ¬ x = y
24 9 13 18 23 syl3anc ⊢ f : ∅ ∅ ⟶ 1-1 A → ∃ x ∈ A ∃ y ∈ A ¬ x = y
25 24 exlimiv ⊢ ∃ f f : ∅ ∅ ⟶ 1-1 A → ∃ x ∈ A ∃ y ∈ A ¬ x = y
26 4 25 syl ⊢ 2 𝑜 ≼ A → ∃ x ∈ A ∃ y ∈ A ¬ x = y