Theorem nf3or 1905

Description: If 𝑥 is not free in 𝜑, 𝜓, and 𝜒, then it is not free in (𝜑 ∨ 𝜓 ∨ 𝜒). (Contributed by Mario Carneiro, 11-Aug-2016.)

Hypotheses

Ref	Expression
nf.1	⊢ Ⅎ𝑥𝜑
nf.2	⊢ Ⅎ𝑥𝜓
nf.3	⊢ Ⅎ𝑥𝜒

Assertion

Ref	Expression
nf3or	⊢ Ⅎ𝑥(𝜑 ∨ 𝜓 ∨ 𝜒)

Proof of Theorem nf3or

Step	Hyp	Ref	Expression
1		df-3or 1088	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒))
2		nf.1	. . . 4 ⊢ Ⅎ𝑥𝜑
3		nf.2	. . . 4 ⊢ Ⅎ𝑥𝜓
4	2, 3	nfor 1904	. . 3 ⊢ Ⅎ𝑥(𝜑 ∨ 𝜓)
5		nf.3	. . 3 ⊢ Ⅎ𝑥𝜒
6	4, 5	nfor 1904	. 2 ⊢ Ⅎ𝑥((𝜑 ∨ 𝜓) ∨ 𝜒)
7	1, 6	nfxfr 1853	1 ⊢ Ⅎ𝑥(𝜑 ∨ 𝜓 ∨ 𝜒)

Syntax hints:   ∨ wo 848   ∨ w3o 1086  Ⅎwnf 1783

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3or 1088  df-ex 1780  df-nf 1784

This theorem is referenced by:  nfso  5599