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Metamath Proof Explorer

Theorem nfntht

Description: Closed form of nfnth . (Contributed by BJ, 16-Sep-2021) (Proof shortened by Wolf Lammen, 4-Sep-2022)

		Ref	Expression
	Assertion	nfntht	$⊢ \neg \exists x φ \to Ⅎ x φ$

Step	Hyp	Ref	Expression
1		pm2.21	$⊢ \neg \exists x φ \to (\exists x φ \to \forall x φ)$
2	1	nfd	$⊢ \neg \exists x φ \to Ⅎ x φ$