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 → ( ∃ 𝑥 𝜓 → ∀ 𝑥 𝜓 ) )
Assertion	nfd	⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 )

Proof

Step	Hyp	Ref	Expression
1		nfd.1	⊢ ( 𝜑 → ( ∃ 𝑥 𝜓 → ∀ 𝑥 𝜓 ) )
2		df-nf	⊢ ( Ⅎ 𝑥 𝜓 ↔ ( ∃ 𝑥 𝜓 → ∀ 𝑥 𝜓 ) )
3	1 2	sylibr	⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 )