Theorem 3orcoma 1090

Description: Commutation law for triple disjunction. (Contributed by Mario Carneiro, 4-Sep-2016.)

Assertion

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

Proof of Theorem 3orcoma

Step	Hyp	Ref	Expression
1		or12 918	. 2 ⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ (𝜓 ∨ (𝜑 ∨ 𝜒)))
2		3orass 1087	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒)))
3		3orass 1087	. 2 ⊢ ((𝜓 ∨ 𝜑 ∨ 𝜒) ↔ (𝜓 ∨ (𝜑 ∨ 𝜒)))
4	1, 2, 3	3bitr4i 302	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜑 ∨ 𝜒))

Syntax hints:   ↔ wb 205   ∨ wo 845   ∨ w3o 1083

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 206  df-or 846  df-3or 1085

This theorem is referenced by:  3orcomb  1091  nogt01o  27675  outpasch  28631  eliccioo  32739