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Metamath Proof Explorer

Theorem nf5di

Description: Since the converse holds by a1i , this inference shows that we can represent a not-free hypothesis with either F/ x ph (inference form) or ( ph -> F/ x ph ) (deduction form). (Contributed by NM, 17-Aug-2018) (Proof shortened by Wolf Lammen, 10-Jul-2019)

		Ref	Expression
	Hypothesis	nf5di.1	\|- ( ph -> F/ x ph )
	Assertion	nf5di	\|- F/ x ph

Proof

Step	Hyp	Ref	Expression
1		nf5di.1	\|- ( ph -> F/ x ph )
2	1	nf5rd	\|- ( ph -> ( ph -> A. x ph ) )
3	2	pm2.43i	\|- ( ph -> A. x ph )
4	3	nf5i	\|- F/ x ph