Theorem 3orcomb 944

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 376	. . 3 ⊢ ((ψ ∨ χ) ↔ (χ ∨ ψ))
2	1	orbi2i 505	. 2 ⊢ ((φ ∨ (ψ ∨ χ)) ↔ (φ ∨ (χ ∨ ψ)))
3		3orass 937	. 2 ⊢ ((φ ∨ ψ ∨ χ) ↔ (φ ∨ (ψ ∨ χ)))
4		3orass 937	. 2 ⊢ ((φ ∨ χ ∨ ψ) ↔ (φ ∨ (χ ∨ ψ)))
5	2, 3, 4	3bitr4i 268	1 ⊢ ((φ ∨ ψ ∨ χ) ↔ (φ ∨ χ ∨ ψ))

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:  eueq3  3012