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Metamath Proof Explorer

Theorem nfsb

Description: If z is not free in ph , then it is not free in [ y / x ] ph when y and z are distinct. See nfsbv for a version with an additional disjoint variable condition on x , z but not requiring ax-13 . (Contributed by Mario Carneiro, 11-Aug-2016) (Proof shortened by Wolf Lammen, 25-Feb-2024) Usage of this theorem is discouraged because it depends on ax-13 . Use nfsbv instead. (New usage is discouraged.)

		Ref	Expression
	Hypothesis	nfsb.1	\|- F/ z ph
	Assertion	nfsb	\|- F/ z [ y / x ] ph

Proof

Step	Hyp	Ref	Expression
1		nfsb.1	\|- F/ z ph
2		nftru	\|- F/ x T.
3	1	a1i	\|- ( T. -> F/ z ph )
4	2 3	nfsbd	\|- ( T. -> F/ z [ y / x ] ph )
5	4	mptru	\|- F/ z [ y / x ] ph