Metamath Proof Explorer


Theorem sbss

Description: Set substitution into the first argument of a subset relation. (Contributed by Rodolfo Medina, 7-Jul-2010) (Proof shortened by Mario Carneiro, 14-Nov-2016)

Ref Expression
Assertion sbss
|- ( [ y / x ] x C_ A <-> y C_ A )

Proof

Step Hyp Ref Expression
1 sseq1
 |-  ( x = z -> ( x C_ A <-> z C_ A ) )
2 sseq1
 |-  ( z = y -> ( z C_ A <-> y C_ A ) )
3 1 2 sbievw2
 |-  ( [ y / x ] x C_ A <-> y C_ A )