Theorem nf3or 1837

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

Hypotheses

Ref	Expression
nf.1	⊢ Ⅎxφ
nf.2	⊢ Ⅎxψ
nf.3	⊢ Ⅎxχ

Assertion

Ref	Expression
nf3or	⊢ Ⅎx(φ ∨ ψ ∨ χ)

Proof of Theorem nf3or

Step	Hyp	Ref	Expression
1		df-3or 935	. 2 ⊢ ((φ ∨ ψ ∨ χ) ↔ ((φ ∨ ψ) ∨ χ))
2		nf.1	. . . 4 ⊢ Ⅎxφ
3		nf.2	. . . 4 ⊢ Ⅎxψ
4	2, 3	nfor 1836	. . 3 ⊢ Ⅎx(φ ∨ ψ)
5		nf.3	. . 3 ⊢ Ⅎxχ
6	4, 5	nfor 1836	. 2 ⊢ Ⅎx((φ ∨ ψ) ∨ χ)
7	1, 6	nfxfr 1570	1 ⊢ Ⅎx(φ ∨ ψ ∨ χ)

This proof depends on syntax axioms:   ∨ wo 357   ∨ w3o 933  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-3or 935  df-tru 1319  df-ex 1542  df-nf 1545

This theorem is used by: (None)