Metamath Proof Explorer


Theorem 3orcoma

Description: Commutation law for triple disjunction. (Contributed by Mario Carneiro, 4-Sep-2016)

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

Proof

Step Hyp Ref Expression
1 or12 ⊢ ( ( 𝜑 ∨ ( 𝜓 ∨ 𝜒 ) ) ↔ ( 𝜓 ∨ ( 𝜑 ∨ 𝜒 ) ) )
2 3orass ⊢ ( ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) ↔ ( 𝜑 ∨ ( 𝜓 ∨ 𝜒 ) ) )
3 3orass ⊢ ( ( 𝜓 ∨ 𝜑 ∨ 𝜒 ) ↔ ( 𝜓 ∨ ( 𝜑 ∨ 𝜒 ) ) )
4 1 2 3 3bitr4i ⊢ ( ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) ↔ ( 𝜓 ∨ 𝜑 ∨ 𝜒 ) )