Metamath Proof Explorer


Theorem sscon34b

Description: Relative complementation reverses inclusion of subclasses. Relativized version of complss . (Contributed by RP, 3-Jun-2021)

Ref Expression
Assertion sscon34b
|- ( ( A C_ C /\ B C_ C ) -> ( A C_ B <-> ( C \ B ) C_ ( C \ A ) ) )

Proof

Step Hyp Ref Expression
1 sscon
 |-  ( A C_ B -> ( C \ B ) C_ ( C \ A ) )
2 sscon
 |-  ( ( C \ B ) C_ ( C \ A ) -> ( C \ ( C \ A ) ) C_ ( C \ ( C \ B ) ) )
3 dfss4
 |-  ( A C_ C <-> ( C \ ( C \ A ) ) = A )
4 3 birani
 |-  ( ( A C_ C /\ B C_ C ) -> ( C \ ( C \ A ) ) = A )
5 dfss4
 |-  ( B C_ C <-> ( C \ ( C \ B ) ) = B )
6 5 bilani
 |-  ( ( A C_ C /\ B C_ C ) -> ( C \ ( C \ B ) ) = B )
7 4 6 sseq12d
 |-  ( ( A C_ C /\ B C_ C ) -> ( ( C \ ( C \ A ) ) C_ ( C \ ( C \ B ) ) <-> A C_ B ) )
8 2 7 imbitrid
 |-  ( ( A C_ C /\ B C_ C ) -> ( ( C \ B ) C_ ( C \ A ) -> A C_ B ) )
9 1 8 impbid2
 |-  ( ( A C_ C /\ B C_ C ) -> ( A C_ B <-> ( C \ B ) C_ ( C \ A ) ) )