Theorem nfor 1466

Description: If 𝑥 is not free in 𝜑 and 𝜓, it is not free in (𝜑 ∨ 𝜓). (Contributed by Jim Kingdon, 11-Mar-2018.)

Hypotheses

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

Assertion

Ref	Expression
nfor	⊢ Ⅎ𝑥(𝜑 ∨ 𝜓)

Proof of Theorem nfor

Step	Hyp	Ref	Expression
1		nfor.1	. . . 4 ⊢ Ⅎ𝑥𝜑
2	1	nfri 1412	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
3		nfor.2	. . . 4 ⊢ Ⅎ𝑥𝜓
4	3	nfri 1412	. . 3 ⊢ (𝜓 → ∀𝑥𝜓)
5	2, 4	hbor 1438	. 2 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))
6	5	nfi 1351	1 ⊢ Ⅎ𝑥(𝜑 ∨ 𝜓)

Syntax hints:   ∨ wo 629  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-io 630  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:  nfdc  1549  nfun  3099  nfpr  3420  nfso  4039  nffrec  5982  indpi  6440  nfsum1  9875  nfsum  9876  bj-findis  10104