Metamath Proof Explorer
Description: Commutation law for triple disjunction. (Contributed by Scott Fenton, 20-Apr-2011) (Proof shortened by Wolf Lammen, 8-Apr-2022)
|
|
Ref |
Expression |
|
Assertion |
3orcomb |
⊢ ( ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) ↔ ( 𝜑 ∨ 𝜒 ∨ 𝜓 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
3orcoma |
⊢ ( ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) ↔ ( 𝜓 ∨ 𝜑 ∨ 𝜒 ) ) |
2 |
|
3orrot |
⊢ ( ( 𝜓 ∨ 𝜑 ∨ 𝜒 ) ↔ ( 𝜑 ∨ 𝜒 ∨ 𝜓 ) ) |
3 |
1 2
|
bitri |
⊢ ( ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) ↔ ( 𝜑 ∨ 𝜒 ∨ 𝜓 ) ) |