Theorem 3orrot 940

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 376	. 2 ⊢ ((φ ∨ (ψ ∨ χ)) ↔ ((ψ ∨ χ) ∨ φ))
2		3orass 937	. 2 ⊢ ((φ ∨ ψ ∨ χ) ↔ (φ ∨ (ψ ∨ χ)))
3		df-3or 935	. 2 ⊢ ((ψ ∨ χ ∨ φ) ↔ ((ψ ∨ χ) ∨ φ))
4	1, 2, 3	3bitr4i 268	1 ⊢ ((φ ∨ ψ ∨ χ) ↔ (ψ ∨ χ ∨ φ))

Syntax hints:   ↔ wb 176   ∨ wo 357   ∨ w3o 933

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 177  df-or 359  df-3or 935

This theorem is referenced by:  3mix2  1125  3mix3  1126  eueq3  3012  tprot  3816  nncdiv3  6278