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  3672  oneltri  6405  soseq  8160  swoso  8734  swrdnd  14724  elnnzs  28662  colcom  28896  legso  28937  lncom  28965  vonf1wev  35690  vonf1owevOLD  35692  colinearperm1  36627  frege129d  44588  ordelordALT  45345  ordelordALTVD  45674  chnerlem3  47697  usgrexmpl2nb3  48935