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Metamath Proof Explorer
Description: Commutation law for triple conjunction. (Contributed by NM, 21-Apr-1994)
(Proof shortened by Wolf Lammen, 5-Jun-2022)
|
|
Ref |
Expression |
|
Assertion |
3ancoma |
⊢ ( ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( 𝜓 ∧ 𝜑 ∧ 𝜒 ) ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
3anan12 |
⊢ ( ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( 𝜓 ∧ ( 𝜑 ∧ 𝜒 ) ) ) |
| 2 |
|
3anass |
⊢ ( ( 𝜓 ∧ 𝜑 ∧ 𝜒 ) ↔ ( 𝜓 ∧ ( 𝜑 ∧ 𝜒 ) ) ) |
| 3 |
1 2
|
bitr4i |
⊢ ( ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( 𝜓 ∧ 𝜑 ∧ 𝜒 ) ) |