Theorem 3orcoma 1109

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 934	. 2 ⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ (𝜓 ∨ (𝜑 ∨ 𝜒)))
2		3orass 1106	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒)))
3		3orass 1106	. 2 ⊢ ((𝜓 ∨ 𝜑 ∨ 𝜒) ↔ (𝜓 ∨ (𝜑 ∨ 𝜒)))
4	1, 2, 3	3bitr4i 306	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜑 ∨ 𝜒))

This proof depends on syntax axioms:   ↔ wb 209   ∨ wo 861   ∨ 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:  3orcomb  1110  3orel2  1515  chnpof1  18708  nogt01o  27911  elzs2  28643  outpasch  29088  eliccioo  33320  usgrexmpl2nb0  48854