Theorem eliminable-abeqv

Description: A theorem used to prove the base case of the Eliminability Theorem (see section comment): abstraction equals variable. (Contributed by BJ, 30-Apr-2024) Beware not to use symmetry of class equality. (Proof modification is discouraged.) (New usage is discouraged.)

	Ref	Expression
Assertion	eliminable-abeqv	|- ( { x | ph } = y <-> A. z ( [ z / x ] ph <-> z e. y ) )

Proof

Step	Hyp	Ref	Expression
1		dfcleq	|- ( { x | ph } = y <-> A. z ( z e. { x | ph } <-> z e. y ) )
2		eliminable-velab	|- ( z e. { x | ph } <-> [ z / x ] ph )
3	2	bibi1i	|- ( ( z e. { x | ph } <-> z e. y ) <-> ( [ z / x ] ph <-> z e. y ) )
4	3	albii	|- ( A. z ( z e. { x | ph } <-> z e. y ) <-> A. z ( [ z / x ] ph <-> z e. y ) )
5	1 4	bitri	|- ( { x | ph } = y <-> A. z ( [ z / x ] ph <-> z e. y ) )