Theorem nfor 1906

Description: If 𝑥 is not free in 𝜑 and 𝜓, then it is not free in (𝜑 ∨ 𝜓). (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 11-Aug-2016.)

Hypotheses

Ref	Expression
nf.1	⊢ Ⅎ𝑥𝜑
nf.2	⊢ Ⅎ𝑥𝜓

Assertion

Ref	Expression
nfor	⊢ Ⅎ𝑥(𝜑 ∨ 𝜓)

Proof of Theorem nfor

Step	Hyp	Ref	Expression
1		df-or 849	. 2 ⊢ ((𝜑 ∨ 𝜓) ↔ (¬ 𝜑 → 𝜓))
2		nf.1	. . . 4 ⊢ Ⅎ𝑥𝜑
3	2	nfn 1859	. . 3 ⊢ Ⅎ𝑥 ¬ 𝜑
4		nf.2	. . 3 ⊢ Ⅎ𝑥𝜓
5	3, 4	nfim 1898	. 2 ⊢ Ⅎ𝑥(¬ 𝜑 → 𝜓)
6	1, 5	nfxfr 1855	1 ⊢ Ⅎ𝑥(𝜑 ∨ 𝜓)

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 848  Ⅎwnf 1785

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-ex 1782  df-nf 1786

This theorem is referenced by:  nf3or  1907  axi12  2706  axbnd  2707  nfun  4110  nfunOLD  4111  nfpr  4636  rabsnifsb  4666  disjxun  5083  fsuppmapnn0fiubex  13954  nfsum1  15652  nfsum  15653  nfcprod1  15873  nfcprod  15874  fdc1  38067  dvdsrabdioph  43238  mnringmulrcld  44655  disjinfi  45622  iundjiun  46888