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 ( CHOICE𝑥 ∈ UFL )
9 1 a1i ( CHOICE𝑥 ∈ V )
10 8 9 2thd ( CHOICE → ( 𝑥 ∈ UFL ↔ 𝑥 ∈ V ) )
11 10 eqrdv ( CHOICE → UFL = V )