Theorem 3orrot 1091

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 867	. 2 ⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ ((𝜓 ∨ 𝜒) ∨ 𝜑))
2		3orass 1089	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒)))
3		df-3or 1087	. 2 ⊢ ((𝜓 ∨ 𝜒 ∨ 𝜑) ↔ ((𝜓 ∨ 𝜒) ∨ 𝜑))
4	1, 2, 3	3bitr4i 303	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜒 ∨ 𝜑))

Syntax hints:   ↔ wb 205   ∨ wo 844   ∨ w3o 1085

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 206  df-or 845  df-3or 1087

This theorem is referenced by:  3orcomb  1093  3mix2  1330  3mix3  1331  3orel2  1484  eueq3  3646  tprot  4685  wemapsolem  9309  ssxr  11044  elnnz  12329  elznn  12335  pfxnd0  14401  colrot1  26920  lnrot1  26984  lnrot2  26985  dfon2lem5  33763  dfon2lem6  33764  nolt02o  33898  nosupbnd2lem1  33918  colinearperm3  34365  wl-exeq  35693  dvasin  35861  frege129d  41371