Theorem 3orrot 1090

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 1088	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒)))
3		df-3or 1086	. 2 ⊢ ((𝜓 ∨ 𝜒 ∨ 𝜑) ↔ ((𝜓 ∨ 𝜒) ∨ 𝜑))
4	1, 2, 3	3bitr4i 302	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜒 ∨ 𝜑))

Syntax hints:   ↔ wb 205   ∨ wo 843   ∨ w3o 1084

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 844  df-3or 1086

This theorem is referenced by:  3orcomb  1092  3mix2  1329  3mix3  1330  eueq3  3641  tprot  4682  wemapsolem  9239  ssxr  10975  elnnz  12259  elznn  12265  pfxnd0  14329  colrot1  26824  lnrot1  26888  lnrot2  26889  3orel2  33556  dfon2lem5  33669  dfon2lem6  33670  nolt02o  33825  nosupbnd2lem1  33845  colinearperm3  34292  wl-exeq  35620  dvasin  35788  frege129d  41260