Theorem nfnan 1904

Description: If 𝑥 is not free in 𝜑 and 𝜓, then it is not free in (𝜑 ⊼ 𝜓). (Contributed by Scott Fenton, 2-Jan-2018.)

Hypotheses

Ref	Expression
nfan.1	⊢ Ⅎ𝑥𝜑
nfan.2	⊢ Ⅎ𝑥𝜓

Assertion

Ref	Expression
nfnan	⊢ Ⅎ𝑥(𝜑 ⊼ 𝜓)

Proof of Theorem nfnan

Step	Hyp	Ref	Expression
1		df-nan 1484	. 2 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
2		nfan.1	. . . 4 ⊢ Ⅎ𝑥𝜑
3		nfan.2	. . . 4 ⊢ Ⅎ𝑥𝜓
4	2, 3	nfan 1903	. . 3 ⊢ Ⅎ𝑥(𝜑 ∧ 𝜓)
5	4	nfn 1861	. 2 ⊢ Ⅎ𝑥 ¬ (𝜑 ∧ 𝜓)
6	1, 5	nfxfr 1856	1 ⊢ Ⅎ𝑥(𝜑 ⊼ 𝜓)

Syntax hints:  ¬ wn 3   ∧ wa 395   ⊼ wnan 1483  Ⅎwnf 1787

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-nan 1484  df-tru 1542  df-ex 1784  df-nf 1788

This theorem is referenced by: (None)