Theorem nfdh 1417

Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Mario Carneiro, 24-Sep-2016.)

Hypotheses

Ref	Expression
nfdh.1	⊢ (𝜑 → ∀𝑥𝜑)
nfdh.2	⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))

Assertion

Ref	Expression
nfdh	⊢ (𝜑 → Ⅎ𝑥𝜓)

Proof of Theorem nfdh

Step	Hyp	Ref	Expression
1		nfdh.1	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
2	1	nfi 1351	. 2 ⊢ Ⅎ𝑥𝜑
3		nfdh.2	. 2 ⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))
4	2, 3	nfd 1416	1 ⊢ (𝜑 → Ⅎ𝑥𝜓)

Syntax hints:   → wi 4  ∀wal 1241  Ⅎwnf 1349

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-4 1400

This theorem depends on definitions:  df-bi 110  df-nf 1350

This theorem is referenced by:  hbsbd  1858