Metamath Proof Explorer


Theorem 3ancoma

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 ) )