Theorem 3orcomb 1110

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 1109	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜑 ∨ 𝜒))
2		3orrot 1108	. 2 ⊢ ((𝜓 ∨ 𝜑 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓))
3	1, 2	bitri 278	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓))

This proof depends on syntax axioms:   ↔ wb 209   ∨ w3o 1102

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-or 861  df-3or 1104

This theorem is used by:  eueq3  3674  oneltri  6404  soseq  8151  swoso  8725  swrdnd  14697  elnnzs  28603  colcom  28836  legso  28877  lncom  28904  vonf1wev  35600  vonf1owevOLD  35602  colinearperm1  36562  frege129d  44517  ordelordALT  45274  ordelordALTVD  45603  chnerlem3  47628  usgrexmpl2nb3  48827