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Metamath Proof Explorer

Theorem sbiev

Description: Conversion of implicit substitution to explicit substitution. Version of sbie with a disjoint variable condition, not requiring ax-13 . See sbievw for a version with a disjoint variable condition requiring fewer axioms. (Contributed by NM, 30-Jun-1994) (Revised by Wolf Lammen, 18-Jan-2023) Remove dependence on ax-10 and shorten proof. (Revised by BJ, 18-Jul-2023) (Proof shortened by SN, 24-Jul-2025)

	Ref	Expression
Hypotheses	sbiev.1	\|- F/ x ps
	sbiev.2	\|- ( x = y -> ( ph <-> ps ) )
Assertion	sbiev	\|- ( [ y / x ] ph <-> ps )

Proof

Step	Hyp	Ref	Expression
1		sbiev.1	\|- F/ x ps
2		sbiev.2	\|- ( x = y -> ( ph <-> ps ) )
3	2	sbbiiev	\|- ( [ y / x ] ph <-> [ y / x ] ps )
4	1	sbf	\|- ( [ y / x ] ps <-> ps )
5	3 4	bitri	\|- ( [ y / x ] ph <-> ps )