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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 ) ) |