Theorem 3orcomb 1100

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 1099	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜑 ∨ 𝜒))
2		3orrot 1098	. 2 ⊢ ((𝜓 ∨ 𝜑 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓))
3	1, 2	bitri 277	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓))

Syntax hints:   ↔ wb 208   ∨ w3o 1092

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 855  df-3or 1094

This theorem is referenced by:  eueq3  3654  oneltri  6357  soseq  8103  swoso  8672  swrdnd  14612  elnnzs  28415  colcom  28648  legso  28689  lncom  28712  vonf1owev  35351  colinearperm1  36305  frege129d  44222  ordelordALT  44996  ordelordALTVD  45325  chnerlem3  47343  usgrexmpl2nb3  48539