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Metamath Proof Explorer

Theorem nfs1f

Description: If x is not free in ph , it is not free in [ y / x ] ph . (Contributed by Mario Carneiro, 11-Aug-2016)

		Ref	Expression
	Hypothesis	nfs1f.1	$⊢ Ⅎ x φ$
	Assertion	nfs1f	$⊢ Ⅎ x [y/ x] φ$

Step	Hyp	Ref	Expression
1		nfs1f.1	$⊢ Ⅎ x φ$
2	1	sbf	$⊢ [y/ x] φ \leftrightarrow φ$
3	2 1	nfxfr	$⊢ Ⅎ x [y/ x] φ$