Theorem bj-sbievw2

Description: Lemma for substitution. (Contributed by BJ, 23-Jul-2023)

	Ref	Expression
Assertion	bj-sbievw2	|- ( [ y / x ] ( ps -> ph ) -> ( ps -> [ y / x ] ph ) )

Proof

Step	Hyp	Ref	Expression
1		sb6	|- ( [ y / x ] ( ps -> ph ) <-> A. x ( x = y -> ( ps -> ph ) ) )
2		bj-sblem2	|- ( A. x ( x = y -> ( ps -> ph ) ) -> ( ( E. x x = y -> ps ) -> A. x ( x = y -> ph ) ) )
3		jarr	|- ( ( ( E. x x = y -> ps ) -> A. x ( x = y -> ph ) ) -> ( ps -> A. x ( x = y -> ph ) ) )
4		sb6	|- ( [ y / x ] ph <-> A. x ( x = y -> ph ) )
5	3 4	syl6ibr	|- ( ( ( E. x x = y -> ps ) -> A. x ( x = y -> ph ) ) -> ( ps -> [ y / x ] ph ) )
6	2 5	syl	|- ( A. x ( x = y -> ( ps -> ph ) ) -> ( ps -> [ y / x ] ph ) )
7	1 6	sylbi	|- ( [ y / x ] ( ps -> ph ) -> ( ps -> [ y / x ] ph ) )