Theorem 3orcomb 1090

Description: Commutation law for triple disjunction. (Contributed by Scott Fenton, 20-Apr-2011.) (Proof shortened by Wolf Lammen, 8-Apr-2022.)

Assertion

Ref	Expression
3orcomb	⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓))

Proof of Theorem 3orcomb

Step	Hyp	Ref	Expression
1		3orcoma 1089	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜑 ∨ 𝜒))
2		3orrot 1088	. 2 ⊢ ((𝜓 ∨ 𝜑 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓))
3	1, 2	bitri 277	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓))

Syntax hints:   ↔ wb 208   ∨ 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:  eueq3  3704  swoso  8324  swrdnd  14018  colcom  26346  legso  26387  lncom  26410  soseq  33098  colinearperm1  33525  frege129d  40115  ordelordALT  40878  ordelordALTVD  41208