Metamath Proof Explorer


Theorem 3orrot

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

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

Proof

Step Hyp Ref Expression
1 orcom ⊢ ( ( 𝜑 ∨ ( 𝜓 ∨ 𝜒 ) ) ↔ ( ( 𝜓 ∨ 𝜒 ) ∨ 𝜑 ) )
2 3orass ⊢ ( ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) ↔ ( 𝜑 ∨ ( 𝜓 ∨ 𝜒 ) ) )
3 df-3or ⊢ ( ( 𝜓 ∨ 𝜒 ∨ 𝜑 ) ↔ ( ( 𝜓 ∨ 𝜒 ) ∨ 𝜑 ) )
4 1 2 3 3bitr4i ⊢ ( ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) ↔ ( 𝜓 ∨ 𝜒 ∨ 𝜑 ) )