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 869	. 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 846   ∨ 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 847  df-3or 1088

This theorem is referenced by:  3orcomb  1094  3mix2  1331  3mix3  1332  3orel2  1484  eueq3  3733  tprot  4774  wemapsolem  9619  ssxr  11359  elnnz  12649  elznn  12655  pfxnd0  14736  nolt02o  27758  nosupbnd2lem1  27778  colrot1  28585  lnrot1  28649  lnrot2  28650  dfon2lem5  35751  dfon2lem6  35752  colinearperm3  36027  wl-exeq  37488  dvasin  37664  frege129d  43725  usgrexmpl2nb0  47846  usgrexmpl2nb3  47849