Theorem isufl 21717

Description: Define the (strong) ultrafilter lemma, parameterized over base sets. A set X satisfies the ultrafilter lemma if every filter on X is a subset of some ultrafilter. (Contributed by Mario Carneiro, 26-Aug-2015.)

Assertion

Ref	Expression
isufl	|- ( X e. V -> ( X e. UFL <-> A. f e. ( Fil ` X ) E. g e. ( UFil ` X ) f C_ g ) )

Distinct variable group:   f, g, X

Allowed substitution hints:   V( f, g)

Proof of Theorem isufl

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		fveq2 6191	. . 3 |- ( x = X -> ( Fil ` x ) = ( Fil ` X ) )
2		fveq2 6191	. . . 4 |- ( x = X -> ( UFil ` x ) = ( UFil ` X ) )
3	2	rexeqdv 3145	. . 3 |- ( x = X -> ( E. g e. ( UFil ` x ) f C_ g <-> E. g e. ( UFil ` X ) f C_ g ) )
4	1, 3	raleqbidv 3152	. 2 |- ( x = X -> ( A. f e. ( Fil ` x ) E. g e. ( UFil ` x ) f C_ g <-> A. f e. ( Fil ` X ) E. g e. ( UFil ` X ) f C_ g ) )
5		df-ufl 21706	. 2 |- UFL = { x | A. f e. ( Fil ` x ) E. g e. ( UFil ` x ) f C_ g }
6	4, 5	elab2g 3353	1 |- ( X e. V -> ( X e. UFL <-> A. f e. ( Fil ` X ) E. g e. ( UFil ` X ) f C_ g ) )

Syntax hints:   -> wi 4   <-> wb 196   = wceq 1483   e. wcel 1990   A.wral 2912   E.wrex 2913   C_ wss 3574   ` cfv 5888   Filcfil 21649   UFilcufil 21703  UFLcufl 21704

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ral 2917  df-rex 2918  df-rab 2921  df-v 3202  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-nul 3916  df-if 4087  df-sn 4178  df-pr 4180  df-op 4184  df-uni 4437  df-br 4654  df-iota 5851  df-fv 5896  df-ufl 21706

This theorem is referenced by:  ufli  21718  numufl  21719  ssufl  21722  ufldom  21766