Metamath Proof Explorer


Theorem acufl

Description: The axiom of choice implies the ultrafilter lemma. (Contributed by Mario Carneiro, 26-Aug-2015)

Ref Expression
Assertion acufl ⊢ CHOICE → UFL = V

Proof

Step Hyp Ref Expression
1 vex ⊢ x ∈ V
2 1 pwex ⊢ 𝒫 x ∈ V
3 2 pwex ⊢ 𝒫 𝒫 x ∈ V
4 dfac10 ⊢ CHOICE ↔ dom ⁡ card = V
5 4 biimpi ⊢ CHOICE → dom ⁡ card = V
6 3 5 eleqtrrid ⊢ CHOICE → 𝒫 𝒫 x ∈ dom ⁡ card
7 numufl ⊢ 𝒫 𝒫 x ∈ dom ⁡ card → x ∈ UFL
8 6 7 syl ⊢ CHOICE → x ∈ UFL
9 1 a1i ⊢ CHOICE → x ∈ V
10 8 9 2thd ⊢ CHOICE → x ∈ UFL ↔ x ∈ V
11 10 eqrdv ⊢ CHOICE → UFL = V