Theorem 3orrot 969

Description: Rotation law for triple disjunction. (Contributed by NM, 4-Apr-1995.)

Assertion

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

Proof of Theorem 3orrot

Step	Hyp	Ref	Expression
1		orcom 718	. 2 ⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ ((𝜓 ∨ 𝜒) ∨ 𝜑))
2		3orass 966	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒)))
3		df-3or 964	. 2 ⊢ ((𝜓 ∨ 𝜒 ∨ 𝜑) ↔ ((𝜓 ∨ 𝜒) ∨ 𝜑))
4	1, 2, 3	3bitr4i 211	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜒 ∨ 𝜑))

Syntax hints:   ↔ wb 104   ∨ wo 698   ∨ w3o 962

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699

This theorem depends on definitions:  df-bi 116  df-3or 964

This theorem is referenced by:  3mix2  1152  3mix3  1153  eueq3dc  2862  tprot  3624  sotritrieq  4255  elnnz  9088  elznn  9094  ztri3or0  9120  zapne  9149