Metamath Proof Explorer


Theorem ssdisj

Description: Intersection with a subclass of a disjoint class. (Contributed by FL, 24-Jan-2007) (Proof shortened by JJ, 14-Jul-2021)

Ref Expression
Assertion ssdisj A B B C = A C =

Proof

Step Hyp Ref Expression
1 ssrin A B A C B C
2 eqimss B C = B C
3 1 2 sylan9ss A B B C = A C
4 ss0 A C A C =
5 3 4 syl A B B C = A C =