Theorem 3orrot 587

Description: Rotation law for triple disjunction.

Assertion

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

Proof of Theorem 3orrot

Step	Hyp	Ref	Expression
1		orcom 209	. 2 ⊢ ((φ ∨ (ψ ∨ χ)) ↔ ((ψ ∨ χ) ∨ φ))
2		3orass 584	. 2 ⊢ ((φ ∨ ψ ∨ χ) ↔ (φ ∨ (ψ ∨ χ)))
3		df-3or 582	. 2 ⊢ ((ψ ∨ χ ∨ φ) ↔ ((ψ ∨ χ) ∨ φ))
4	1, 2, 3	3bitr4 158	1 ⊢ ((φ ∨ ψ ∨ χ) ↔ (ψ ∨ χ ∨ φ))

Syntax hints:   ↔ wb 127   ∨ wo 195   ∨ w3o 580

This theorem is referenced by:  3mix2 601  3mix3 602  elnnz 4572  elnnz1 4581

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-or 197  df-3or 582

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