Metamath Proof Explorer

Theorem sbthlem3

Description: Lemma for sbth . (Contributed by NM, 22-Mar-1998)

Ref Expression
Hypotheses sbthlem.1 
|- A e. _V
sbthlem.2 
|- D = { x | ( x C_ A /\ ( g " ( B \ ( f " x ) ) ) C_ ( A \ x ) ) }
Assertion sbthlem3 
|- ( ran g C_ A -> ( g " ( B \ ( f " U. D ) ) ) = ( A \ U. D ) )

Proof

Step Hyp Ref Expression
1 sbthlem.1 
 |-  A e. _V
2 sbthlem.2 
 |-  D = { x | ( x C_ A /\ ( g " ( B \ ( f " x ) ) ) C_ ( A \ x ) ) }
3 1 2 sbthlem2 
 |-  ( ran g C_ A -> ( A \ ( g " ( B \ ( f " U. D ) ) ) ) C_ U. D )
4 1 2 sbthlem1 
 |-  U. D C_ ( A \ ( g " ( B \ ( f " U. D ) ) ) )
5 3 4 jctil 
 |-  ( ran g C_ A -> ( U. D C_ ( A \ ( g " ( B \ ( f " U. D ) ) ) ) /\ ( A \ ( g " ( B \ ( f " U. D ) ) ) ) C_ U. D ) )
6 eqss 
 |-  ( U. D = ( A \ ( g " ( B \ ( f " U. D ) ) ) ) <-> ( U. D C_ ( A \ ( g " ( B \ ( f " U. D ) ) ) ) /\ ( A \ ( g " ( B \ ( f " U. D ) ) ) ) C_ U. D ) )
7 5 6 sylibr 
 |-  ( ran g C_ A -> U. D = ( A \ ( g " ( B \ ( f " U. D ) ) ) ) )
8 7 difeq2d 
 |-  ( ran g C_ A -> ( A \ U. D ) = ( A \ ( A \ ( g " ( B \ ( f " U. D ) ) ) ) ) )
9 imassrn 
 |-  ( g " ( B \ ( f " U. D ) ) ) C_ ran g
10 sstr2 
 |-  ( ( g " ( B \ ( f " U. D ) ) ) C_ ran g -> ( ran g C_ A -> ( g " ( B \ ( f " U. D ) ) ) C_ A ) )
11 9 10 ax-mp 
 |-  ( ran g C_ A -> ( g " ( B \ ( f " U. D ) ) ) C_ A )
12 dfss4 
 |-  ( ( g " ( B \ ( f " U. D ) ) ) C_ A <-> ( A \ ( A \ ( g " ( B \ ( f " U. D ) ) ) ) ) = ( g " ( B \ ( f " U. D ) ) ) )
13 11 12 sylib 
 |-  ( ran g C_ A -> ( A \ ( A \ ( g " ( B \ ( f " U. D ) ) ) ) ) = ( g " ( B \ ( f " U. D ) ) ) )
14 8 13 eqtr2d 
 |-  ( ran g C_ A -> ( g " ( B \ ( f " U. D ) ) ) = ( A \ U. D ) )