Theorem nfd

Description: Deduce that x is not free in ps in a context. (Contributed by Wolf Lammen, 16-Sep-2021)

	Ref	Expression
Hypothesis	nfd.1	|- ( ph -> ( E. x ps -> A. x ps ) )
Assertion	nfd	|- ( ph -> F/ x ps )

Proof

Step	Hyp	Ref	Expression
1		nfd.1	|- ( ph -> ( E. x ps -> A. x ps ) )
2		df-nf	|- ( F/ x ps <-> ( E. x ps -> A. x ps ) )
3	1 2	sylibr	|- ( ph -> F/ x ps )