Theorem sbcbi2

Description: Substituting into equivalent wff's gives equivalent results. (Contributed by Giovanni Mascellani, 9-Apr-2018) (Proof shortened by Wolf Lammen, 4-May-2023) Avoid ax-10, ax-12. (Revised by Steven Nguyen, 5-May-2024)

	Ref	Expression
Assertion	sbcbi2	|- ( A. x ( ph <-> ps ) -> ( [. A / x ]. ph <-> [. A / x ]. ps ) )

Proof

Step	Hyp	Ref	Expression
1		abbi1	|- ( A. x ( ph <-> ps ) -> { x | ph } = { x | ps } )
2		eleq2	|- ( { x | ph } = { x | ps } -> ( A e. { x | ph } <-> A e. { x | ps } ) )
3	1 2	syl	|- ( A. x ( ph <-> ps ) -> ( A e. { x | ph } <-> A e. { x | ps } ) )
4		df-sbc	|- ( [. A / x ]. ph <-> A e. { x | ph } )
5		df-sbc	|- ( [. A / x ]. ps <-> A e. { x | ps } )
6	3 4 5	3bitr4g	|- ( A. x ( ph <-> ps ) -> ( [. A / x ]. ph <-> [. A / x ]. ps ) )