Theorem 3orcoma 1093

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 921	. 2 ⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ (𝜓 ∨ (𝜑 ∨ 𝜒)))
2		3orass 1090	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒)))
3		3orass 1090	. 2 ⊢ ((𝜓 ∨ 𝜑 ∨ 𝜒) ↔ (𝜓 ∨ (𝜑 ∨ 𝜒)))
4	1, 2, 3	3bitr4i 303	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜑 ∨ 𝜒))

Syntax hints:   ↔ wb 206   ∨ wo 848   ∨ w3o 1086

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 207  df-or 849  df-3or 1088

This theorem is referenced by:  3orcomb  1094  3orel2  1487  chnpof1  18565  nogt01o  27676  elzs2  28407  outpasch  28839  eliccioo  33023  usgrexmpl2nb0  48391