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  2707  axbnd  2708  nfun  4111  nfunOLD  4112  nfpr  4637  rabsnifsb  4667  disjxun  5084  fsuppmapnn0fiubex  13945  nfsum1  15643  nfsum  15644  nfcprod1  15864  nfcprod  15865  fdc1  38081  dvdsrabdioph  43256  mnringmulrcld  44673  disjinfi  45640  iundjiun  46906