Theorem 3orcoma 942

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 509	. 2 ⊢ ((φ ∨ (ψ ∨ χ)) ↔ (ψ ∨ (φ ∨ χ)))
2		3orass 937	. 2 ⊢ ((φ ∨ ψ ∨ χ) ↔ (φ ∨ (ψ ∨ χ)))
3		3orass 937	. 2 ⊢ ((ψ ∨ φ ∨ χ) ↔ (ψ ∨ (φ ∨ χ)))
4	1, 2, 3	3bitr4i 268	1 ⊢ ((φ ∨ ψ ∨ χ) ↔ (ψ ∨ φ ∨ χ))

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:  cadcomb  1396