Theorem nfor 1627

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 1572	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
3		nfor.2	. . . 4 ⊢ Ⅎ𝑥𝜓
4	3	nfri 1572	. . 3 ⊢ (𝜓 → ∀𝑥𝜓)
5	2, 4	hbor 1599	. 2 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))
6	5	nfi 1515	1 ⊢ Ⅎ𝑥(𝜑 ∨ 𝜓)

This proof depends on syntax axioms:   ∨ wo 720  Ⅎwnf 1513

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-gen 1502  ax-4 1563

This proof depends on definitions:  df-bi 117  df-nf 1514

This theorem is used by:  nfdc  1711  nfun  3385  nfpr  3759  rabsnifsb  3777  nfso  4447  nffrec  6667  indpi  7709  nfsum1  12122  nfsum  12123  nfcprod1  12321  nfcprod  12322  bj-findis  17005  isomninnlem  17079  trirec0  17093