Theorem nf5d 2325

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

Hypotheses

Ref	Expression
nf5d.1	⊢ Ⅎ𝑥𝜑
nf5d.2	⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))

Assertion

Ref	Expression
nf5d	⊢ (𝜑 → Ⅎ𝑥𝜓)

Proof of Theorem nf5d

Step	Hyp	Ref	Expression
1		nf5d.1	. . 3 ⊢ Ⅎ𝑥𝜑
2		nf5d.2	. . 3 ⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))
3	1, 2	alrimi 2255	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 → ∀𝑥𝜓))
4		nf5-1 2186	. 2 ⊢ (∀𝑥(𝜓 → ∀𝑥𝜓) → Ⅎ𝑥𝜓)
5	3, 4	syl 18	1 ⊢ (𝜑 → Ⅎ𝑥𝜓)

Syntax hints:   → wi 4  ∀wal 1565  Ⅎwnf 1810

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-10 2182  ax-12 2219

This theorem depends on definitions:  df-bi 210  df-ex 1807  df-nf 1811

This theorem is referenced by:  dvelimhw  2383  nfeqf  2419  cbv1h  2443  axc16nfALT  2475  nfsb2  2521  distel  36192  mh-setindnd  36937  bj-cbv1hv  37320  ichnfimlem  48101