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Metamath Proof Explorer

Theorem nf5dh

Description: Deduce that x is not free in ps in a context. (Contributed by Mario Carneiro, 24-Sep-2016) df-nf changed. (Revised by Wolf Lammen, 11-Oct-2021)

	Ref	Expression
Hypotheses	nf5dh.1	$⊢ φ \to \forall x φ$
	nf5dh.2	$⊢ φ \to (ψ \to \forall x ψ)$
Assertion	nf5dh	$⊢ φ \to Ⅎ x ψ$

Proof

Step	Hyp	Ref	Expression
1		nf5dh.1	$⊢ φ \to \forall x φ$
2		nf5dh.2	$⊢ φ \to (ψ \to \forall x ψ)$
3	1 2	alrimih	$⊢ φ \to \forall x (ψ \to \forall x ψ)$
4		nf5-1	$⊢ \forall x (ψ \to \forall x ψ) \to Ⅎ x ψ$
5	3 4	syl	$⊢ φ \to Ⅎ x ψ$