Theorem nfor 1836

Description: If x is not free in φ and ψ, it is not free in (φ ∨ ψ). (Contributed 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 359	. 2 ⊢ ((φ ∨ ψ) ↔ (¬ φ → ψ))
2		nf.1	. . . 4 ⊢ Ⅎxφ
3	2	nfn 1793	. . 3 ⊢ Ⅎx ¬ φ
4		nf.2	. . 3 ⊢ Ⅎxψ
5	3, 4	nfim 1813	. 2 ⊢ Ⅎx(¬ φ → ψ)
6	1, 5	nfxfr 1570	1 ⊢ Ⅎx(φ ∨ ψ)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∨ wo 357  Ⅎwnf 1544

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-11 1746

This proof depends on definitions:  df-bi 177  df-or 359  df-tru 1319  df-ex 1542  df-nf 1545

This theorem is used by:  nf3or  1837  axi12  2333  nfpr  3774