Theorem nf5dh

Description: Deduce that x is not free in ps in a context. (Contributed by Mario Carneiro, 24-Sep-2016) df-nf changed. (Revised by Wolf Lammen, 11-Oct-2021)

	Ref	Expression
Hypotheses	nf5dh.1	|- ( ph -> A. x ph )
	nf5dh.2	|- ( ph -> ( ps -> A. x ps ) )
Assertion	nf5dh	|- ( ph -> F/ x ps )

Proof

Step	Hyp	Ref	Expression
1		nf5dh.1	|- ( ph -> A. x ph )
2		nf5dh.2	|- ( ph -> ( ps -> A. x ps ) )
3	1 2	alrimih	|- ( ph -> A. x ( ps -> A. x ps ) )
4		nf5-1	|- ( A. x ( ps -> A. x ps ) -> F/ x ps )
5	3 4	syl	|- ( ph -> F/ x ps )