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Metamath Proof Explorer

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	$⊢ φ \to (\exists x ψ \to \forall x ψ)$
	Assertion	nfd	$⊢ φ \to Ⅎ x ψ$

Step	Hyp	Ref	Expression
1		nfd.1	$⊢ φ \to (\exists x ψ \to \forall x ψ)$
2		df-nf	$⊢ Ⅎ x ψ \leftrightarrow (\exists x ψ \to \forall x ψ)$
3	1 2	sylibr	$⊢ φ \to Ⅎ x ψ$