Metamath Proof Explorer


Theorem bj-sscon

Description: Contraposition law for relative subclasses. Relative and generalized version of ssconb , which it can shorten, as well as conss2 . (Contributed by BJ, 11-Nov-2021) This proof does not rely, even indirectly, on ssconb nor conss2 . (Proof modification is discouraged.)

Ref Expression
Assertion bj-sscon
|- ( ( A i^i V ) C_ ( V \ B ) <-> ( B i^i V ) C_ ( V \ A ) )

Proof

Step Hyp Ref Expression
1 incom
 |-  ( A i^i B ) = ( B i^i A )
2 1 ineq1i
 |-  ( ( A i^i B ) i^i V ) = ( ( B i^i A ) i^i V )
3 2 eqeq1i
 |-  ( ( ( A i^i B ) i^i V ) = (/) <-> ( ( B i^i A ) i^i V ) = (/) )
4 bj-disj2r
 |-  ( ( A i^i V ) C_ ( V \ B ) <-> ( ( A i^i B ) i^i V ) = (/) )
5 bj-disj2r
 |-  ( ( B i^i V ) C_ ( V \ A ) <-> ( ( B i^i A ) i^i V ) = (/) )
6 3 4 5 3bitr4i
 |-  ( ( A i^i V ) C_ ( V \ B ) <-> ( B i^i V ) C_ ( V \ A ) )