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  3657  tprot  4693  wemapsolem  9465  ssxr  11215  elnnz  12534  elznn  12540  pfxnd0  14651  nolt02o  27659  nosupbnd2lem1  27679  colrot1  28627  lnrot1  28691  lnrot2  28692  dfon2lem5  35967  dfon2lem6  35968  colinearperm3  36245  wl-exeq  37859  dvasin  38025  frege129d  44190  usgrexmpl2nb0  48507  usgrexmpl2nb3  48510