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Metamath Proof Explorer

Theorem nfsbv

Description: If z is not free in ph , then it is not free in [ y / x ] ph when z is disjoint from both x and y . Version of nfsb with an additional disjoint variable condition on x , z but not requiring ax-13 . (Contributed by Mario Carneiro, 11-Aug-2016) (Revised by Wolf Lammen, 7-Feb-2023) Remove disjoint variable condition on x , y . (Revised by Steven Nguyen, 13-Aug-2023) (Proof shortened by Wolf Lammen, 25-Oct-2024)

		Ref	Expression
	Hypothesis	nfsbv.nf	\|- F/ z ph
	Assertion	nfsbv	\|- F/ z [ y / x ] ph

Proof

Step	Hyp	Ref	Expression
1		nfsbv.nf	\|- F/ z ph
2	1	nf5ri	\|- ( ph -> A. z ph )
3	2	hbsbw	\|- ( [ y / x ] ph -> A. z [ y / x ] ph )
4	3	nf5i	\|- F/ z [ y / x ] ph