Metamath Proof Explorer


Theorem fiufl

Description: A finite set satisfies the ultrafilter lemma. (Contributed by Mario Carneiro, 26-Aug-2015)

Ref Expression
Assertion fiufl
|- ( X e. Fin -> X e. UFL )

Proof

Step Hyp Ref Expression
1 pwfi
 |-  ( X e. Fin <-> ~P X e. Fin )
2 pwfi
 |-  ( ~P X e. Fin <-> ~P ~P X e. Fin )
3 finnum
 |-  ( ~P ~P X e. Fin -> ~P ~P X e. dom card )
4 numufl
 |-  ( ~P ~P X e. dom card -> X e. UFL )
5 3 4 syl
 |-  ( ~P ~P X e. Fin -> X e. UFL )
6 2 5 sylbi
 |-  ( ~P X e. Fin -> X e. UFL )
7 1 6 sylbi
 |-  ( X e. Fin -> X e. UFL )