Theorem eliminable-veqab

Description: A theorem used to prove the base case of the Eliminability Theorem (see section comment): variable equals abstraction. (Contributed by BJ, 30-Apr-2024) (Proof modification is discouraged.) (New usage is discouraged.)

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

Proof

Step	Hyp	Ref	Expression
1		dfcleq	|- ( x = { y | ph } <-> A. z ( z e. x <-> z e. { y | ph } ) )
2		eliminable-velab	|- ( z e. { y | ph } <-> [ z / y ] ph )
3	2	bibi2i	|- ( ( z e. x <-> z e. { y | ph } ) <-> ( z e. x <-> [ z / y ] ph ) )
4	3	albii	|- ( A. z ( z e. x <-> z e. { y | ph } ) <-> A. z ( z e. x <-> [ z / y ] ph ) )
5	1 4	bitri	|- ( x = { y | ph } <-> A. z ( z e. x <-> [ z / y ] ph ) )