Theorem 3orrot 1092

Description: Rotation law for triple disjunction. (Contributed by NM, 4-Apr-1995.)

Assertion

Ref	Expression
3orrot	⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜒 ∨ 𝜑))

Proof of Theorem 3orrot

Step	Hyp	Ref	Expression
1		orcom 871	. 2 ⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ ((𝜓 ∨ 𝜒) ∨ 𝜑))
2		3orass 1090	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒)))
3		df-3or 1088	. 2 ⊢ ((𝜓 ∨ 𝜒 ∨ 𝜑) ↔ ((𝜓 ∨ 𝜒) ∨ 𝜑))
4	1, 2, 3	3bitr4i 303	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜒 ∨ 𝜑))

Syntax hints:   ↔ wb 206   ∨ wo 848   ∨ w3o 1086

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

This theorem depends on definitions:  df-bi 207  df-or 849  df-3or 1088

This theorem is referenced by:  3orcomb  1094  3mix2  1333  3mix3  1334  3orel2OLD  1488  eueq3  3671  tprot  4708  wemapsolem  9467  ssxr  11214  elnnz  12510  elznn  12516  pfxnd0  14624  nolt02o  27675  nosupbnd2lem1  27695  colrot1  28643  lnrot1  28707  lnrot2  28708  dfon2lem5  36001  dfon2lem6  36002  colinearperm3  36279  wl-exeq  37789  dvasin  37955  frege129d  44119  usgrexmpl2nb0  48391  usgrexmpl2nb3  48394