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	$$⊢\phi \to \forall x\phi$$
	nf5dh.2	$$⊢\phi \to \left(\psi \to \forall x\psi \right)$$
Assertion	nf5dh	$$⊢\phi \to Ⅎx\psi$$

Proof

Step	Hyp	Ref	Expression
1		nf5dh.1	$$⊢\phi \to \forall x\phi$$
2		nf5dh.2	$$⊢\phi \to \left(\psi \to \forall x\psi \right)$$
3	1 2	alrimih	$$⊢\phi \to \forall x\left(\psi \to \forall x\psi \right)$$
4		nf5-1	$$⊢\forall x\left(\psi \to \forall x\psi \right)\to Ⅎx\psi$$
5	3 4	syl	$$⊢\phi \to Ⅎx\psi$$