Theorem nfor 1598

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 1543	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
3		nfor.2	. . . 4 ⊢ Ⅎ𝑥𝜓
4	3	nfri 1543	. . 3 ⊢ (𝜓 → ∀𝑥𝜓)
5	2, 4	hbor 1570	. 2 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))
6	5	nfi 1486	1 ⊢ Ⅎ𝑥(𝜑 ∨ 𝜓)

Syntax hints:   ∨ wo 710  Ⅎwnf 1484

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 711  ax-5 1471  ax-gen 1473  ax-4 1534

This theorem depends on definitions:  df-bi 117  df-nf 1485

This theorem is referenced by:  nfdc  1683  nfun  3333  nfpr  3688  nfso  4357  nffrec  6495  indpi  7475  nfsum1  11742  nfsum  11743  nfcprod1  11940  nfcprod  11941  bj-findis  16053  isomninnlem  16110  trirec0  16124