Theorem abbib

Description: Equal class abstractions require equivalent formulas, and conversely. (Contributed by NM, 25-Nov-2013) (Revised by Mario Carneiro, 11-Aug-2016) Remove dependency on ax-8 and df-clel (by avoiding use of cleqh ). (Revised by BJ, 23-Jun-2019) Reverse the biconditional. (Revised by Wolf Lammen, 23-Feb-2025) (Proof shortened by SN, 22-Oct-2026)

	Ref	Expression
Assertion	abbib	|- ( { x | ph } = { x | ps } <-> A. x ( ph <-> ps ) )

Proof

Step	Hyp	Ref	Expression
1		df-clab	|- ( y e. { x | ph } <-> [ y / x ] ph )
2		df-clab	|- ( y e. { x | ps } <-> [ y / x ] ps )
3	1 2	bibi12i	|- ( ( y e. { x | ph } <-> y e. { x | ps } ) <-> ( [ y / x ] ph <-> [ y / x ] ps ) )
4		sbbi	|- ( [ y / x ] ( ph <-> ps ) <-> ( [ y / x ] ph <-> [ y / x ] ps ) )
5	3 4	bitr4i	|- ( ( y e. { x | ph } <-> y e. { x | ps } ) <-> [ y / x ] ( ph <-> ps ) )
6	5	albii	|- ( A. y ( y e. { x | ph } <-> y e. { x | ps } ) <-> A. y [ y / x ] ( ph <-> ps ) )
7		dfcleq	|- ( { x | ph } = { x | ps } <-> A. y ( y e. { x | ph } <-> y e. { x | ps } ) )
8		sb8v	|- ( A. x ( ph <-> ps ) <-> A. y [ y / x ] ( ph <-> ps ) )
9	6 7 8	3bitr4i	|- ( { x | ph } = { x | ps } <-> A. x ( ph <-> ps ) )