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 → ∀ 𝑥 𝜑 )
	nf5dh.2	⊢ ( 𝜑 → ( 𝜓 → ∀ 𝑥 𝜓 ) )
Assertion	nf5dh	⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 )

Proof

Step	Hyp	Ref	Expression
1		nf5dh.1	⊢ ( 𝜑 → ∀ 𝑥 𝜑 )
2		nf5dh.2	⊢ ( 𝜑 → ( 𝜓 → ∀ 𝑥 𝜓 ) )
3	1 2	alrimih	⊢ ( 𝜑 → ∀ 𝑥 ( 𝜓 → ∀ 𝑥 𝜓 ) )
4		nf5-1	⊢ ( ∀ 𝑥 ( 𝜓 → ∀ 𝑥 𝜓 ) → Ⅎ 𝑥 𝜓 )
5	3 4	syl	⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 )