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 870	. 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 206   ∨ wo 847   ∨ 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 207  df-or 848  df-3or 1087

This theorem is referenced by:  3orcomb  1093  3mix2  1332  3mix3  1333  3orel2OLD  1487  eueq3  3685  tprot  4716  wemapsolem  9510  ssxr  11250  elnnz  12546  elznn  12552  pfxnd0  14660  nolt02o  27614  nosupbnd2lem1  27634  colrot1  28493  lnrot1  28557  lnrot2  28558  dfon2lem5  35782  dfon2lem6  35783  colinearperm3  36058  wl-exeq  37529  dvasin  37705  frege129d  43759  usgrexmpl2nb0  48026  usgrexmpl2nb3  48029