Theorem 3orcomb 989

Description: Commutation law for triple disjunction. (Contributed by Scott Fenton, 20-Apr-2011.)

Assertion

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

Proof of Theorem 3orcomb

Step	Hyp	Ref	Expression
1		orcom 729	. . 3 ⊢ ((𝜓 ∨ 𝜒) ↔ (𝜒 ∨ 𝜓))
2	1	orbi2i 763	. 2 ⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ (𝜑 ∨ (𝜒 ∨ 𝜓)))
3		3orass 983	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒)))
4		3orass 983	. 2 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜓) ↔ (𝜑 ∨ (𝜒 ∨ 𝜓)))
5	2, 3, 4	3bitr4i 212	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓))

Syntax hints:   ↔ wb 105   ∨ wo 709   ∨ w3o 979

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710

This theorem depends on definitions:  df-bi 117  df-3or 981

This theorem is referenced by:  eueq3dc  2938  sotritrieq  4360  exmidontriimlem3  7290