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Metamath Proof Explorer

Theorem nf5i

Description: Deduce that x is not free in ph from the definition. (Contributed by Mario Carneiro, 11-Aug-2016)

		Ref	Expression
	Hypothesis	nf5i.1	$⊢ φ \to \forall x φ$
	Assertion	nf5i	$⊢ Ⅎ x φ$

Step	Hyp	Ref	Expression
1		nf5i.1	$⊢ φ \to \forall x φ$
2		nf5-1	$⊢ \forall x (φ \to \forall x φ) \to Ⅎ x φ$
3	2 1	mpg	$⊢ Ⅎ x φ$