Theorem nfor 1562

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 1507	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
3		nfor.2	. . . 4 ⊢ Ⅎ𝑥𝜓
4	3	nfri 1507	. . 3 ⊢ (𝜓 → ∀𝑥𝜓)
5	2, 4	hbor 1534	. 2 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))
6	5	nfi 1450	1 ⊢ Ⅎ𝑥(𝜑 ∨ 𝜓)

Syntax hints:   ∨ wo 698  Ⅎwnf 1448

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1435  ax-gen 1437  ax-4 1498

This theorem depends on definitions:  df-bi 116  df-nf 1449

This theorem is referenced by:  nfdc  1647  nfun  3278  nfpr  3626  nfso  4280  nffrec  6364  indpi  7283  nfsum1  11297  nfsum  11298  nfcprod1  11495  nfcprod  11496  bj-findis  13861  isomninnlem  13909  trirec0  13923