Theorem nfor 1463

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 1409	. . 3 ⊢ (φ → ∀xφ)
3		nfor.2	. . . 4 ⊢ Ⅎxψ
4	3	nfri 1409	. . 3 ⊢ (ψ → ∀xψ)
5	2, 4	hbor 1435	. 2 ⊢ ((φ ∨ ψ) → ∀x(φ ∨ ψ))
6	5	nfi 1348	1 ⊢ Ⅎx(φ ∨ ψ)

Syntax hints:   ∨ wo 628  Ⅎwnf 1346

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-gen 1335  ax-4 1397

This theorem depends on definitions:  df-bi 110  df-nf 1347

This theorem is referenced by:  nfdc  1546  nfun  3093  nfpr  3411  nfso  4030  nffrec  5921  indpi  6326  bj-findis  9439