Theorem bj-sbel1

Description: Version of sbcel1g when substituting a set. (Note: one could have a corresponding version of sbcel12 when substituting a set, but the point here is that the antecedent of sbcel1g is not needed when substituting a set.) (Contributed by BJ, 6-Oct-2018)

	Ref	Expression
Assertion	bj-sbel1	|- ( [ y / x ] A e. B <-> [_ y / x ]_ A e. B )

Proof

Step	Hyp	Ref	Expression
1		sbsbc	|- ( [ y / x ] A e. B <-> [. y / x ]. A e. B )
2		sbcel1g	|- ( y e. _V -> ( [. y / x ]. A e. B <-> [_ y / x ]_ A e. B ) )
3	2	elv	|- ( [. y / x ]. A e. B <-> [_ y / x ]_ A e. B )
4	1 3	bitri	|- ( [ y / x ] A e. B <-> [_ y / x ]_ A e. B )