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Metamath Proof Explorer

Theorem cbvabv

Description: Rule used to change bound variables, using implicit substitution. Version of cbvab with disjoint variable conditions requiring fewer axioms. (Contributed by NM, 26-May-1999) Require x , y be disjoint to avoid ax-11 and ax-13 . (Revised by Steven Nguyen, 4-Dec-2022)

		Ref	Expression
	Hypothesis	cbvabv.1	\|- ( x = y -> ( ph <-> ps ) )
	Assertion	cbvabv	\|- { x \| ph } = { y \| ps }

Proof

Step	Hyp	Ref	Expression
1		cbvabv.1	\|- ( x = y -> ( ph <-> ps ) )
2	1	cbvsbv	\|- ( [ z / x ] ph <-> [ z / y ] ps )
3		df-clab	\|- ( z e. { x \| ph } <-> [ z / x ] ph )
4		df-clab	\|- ( z e. { y \| ps } <-> [ z / y ] ps )
5	2 3 4	3bitr4i	\|- ( z e. { x \| ph } <-> z e. { y \| ps } )
6	5	eqriv	\|- { x \| ph } = { y \| ps }