Metamath Proof Explorer

Theorem sseq2

Description: Equality theorem for the subclass relationship. (Contributed by NM, 25-Jun-1998)

Ref Expression
Assertion sseq2 
|- ( A = B -> ( C C_ A <-> C C_ B ) )

Proof

Step Hyp Ref Expression
1 eqss 
 |-  ( A = B <-> ( A C_ B /\ B C_ A ) )
2 sstr2 
 |-  ( C C_ A -> ( A C_ B -> C C_ B ) )
3 2 com12 
 |-  ( A C_ B -> ( C C_ A -> C C_ B ) )
4 sstr2 
 |-  ( C C_ B -> ( B C_ A -> C C_ A ) )
5 4 com12 
 |-  ( B C_ A -> ( C C_ B -> C C_ A ) )
6 3 5 anbiim 
 |-  ( ( A C_ B /\ B C_ A ) -> ( C C_ A <-> C C_ B ) )
7 1 6 sylbi 
 |-  ( A = B -> ( C C_ A <-> C C_ B ) )