Theorem nfor 1623

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 1568	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
3		nfor.2	. . . 4 ⊢ Ⅎ𝑥𝜓
4	3	nfri 1568	. . 3 ⊢ (𝜓 → ∀𝑥𝜓)
5	2, 4	hbor 1595	. 2 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))
6	5	nfi 1511	1 ⊢ Ⅎ𝑥(𝜑 ∨ 𝜓)

Syntax hints:   ∨ wo 716  Ⅎwnf 1509

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 717  ax-5 1496  ax-gen 1498  ax-4 1559

This theorem depends on definitions:  df-bi 117  df-nf 1510

This theorem is referenced by:  nfdc  1707  nfun  3365  nfpr  3723  rabsnifsb  3741  nfso  4405  nffrec  6605  indpi  7605  nfsum1  11979  nfsum  11980  nfcprod1  12178  nfcprod  12179  bj-findis  16678  isomninnlem  16745  trirec0  16759