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 862  df-3or 1104

This theorem is used by:  eueq3  3668  oneltri  6406  soseq  8176  swoso  8752  swrdnd  14804  elnnzs  28787  colcom  29021  legso  29062  lncom  29090  vonf1wev  35887  vonf1owevOLD  35889  colinearperm1  36827  frege129d  44763  ordelordALT  45520  ordelordALTVD  45849  chnerlem3  47894  usgrexmpl2nb3  49132