Theorem eqabbw

Description: Version of eqabb using implicit substitution, which requires fewer axioms. (Contributed by GG and AV, 18-Sep-2024)

	Ref	Expression
Hypothesis	eqabbw.1	|- ( x = y -> ( ph <-> ps ) )
Assertion	eqabbw	|- ( A = { x | ph } <-> A. y ( y e. A <-> ps ) )

Proof

Step	Hyp	Ref	Expression
1		eqabbw.1	|- ( x = y -> ( ph <-> ps ) )
2		dfcleq	|- ( A = { x | ph } <-> A. y ( y e. A <-> y e. { x | ph } ) )
3		df-clab	|- ( y e. { x | ph } <-> [ y / x ] ph )
4	1	sbievw	|- ( [ y / x ] ph <-> ps )
5	3 4	bitri	|- ( y e. { x | ph } <-> ps )
6	5	bibi2i	|- ( ( y e. A <-> y e. { x | ph } ) <-> ( y e. A <-> ps ) )
7	6	albii	|- ( A. y ( y e. A <-> y e. { x | ph } ) <-> A. y ( y e. A <-> ps ) )
8	2 7	bitri	|- ( A = { x | ph } <-> A. y ( y e. A <-> ps ) )