Theorem 3orrot 1088

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 866	. 2 ⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ ((𝜓 ∨ 𝜒) ∨ 𝜑))
2		3orass 1086	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒)))
3		df-3or 1084	. 2 ⊢ ((𝜓 ∨ 𝜒 ∨ 𝜑) ↔ ((𝜓 ∨ 𝜒) ∨ 𝜑))
4	1, 2, 3	3bitr4i 305	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜒 ∨ 𝜑))

Syntax hints:   ↔ wb 208   ∨ wo 843   ∨ w3o 1082

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 209  df-or 844  df-3or 1084

This theorem is referenced by:  3orcomb  1090  3mix2  1327  3mix3  1328  eueq3  3702  tprot  4679  wemapsolem  9008  ssxr  10704  elnnz  11985  elznn  11991  pfxnd0  14044  colrot1  26339  lnrot1  26403  lnrot2  26404  3orel2  32936  dfon2lem5  33027  dfon2lem6  33028  nolt02o  33194  nosupbnd2lem1  33210  colinearperm3  33519  wl-exeq  34768  dvasin  34972  frege129d  40101