Metamath Proof Explorer


Theorem andir

Description: Distributive law for conjunction. (Contributed by NM, 12-Aug-1994)

Ref Expression
Assertion andir ( ( ( 𝜑 ∨ 𝜓 ) ∧ 𝜒 ) ↔ ( ( 𝜑 ∧ 𝜒 ) ∨ ( 𝜓 ∧ 𝜒 ) ) )

Proof

Step Hyp Ref Expression
1 andi ⊢ ( ( 𝜒 ∧ ( 𝜑 ∨ 𝜓 ) ) ↔ ( ( 𝜒 ∧ 𝜑 ) ∨ ( 𝜒 ∧ 𝜓 ) ) )
2 ancom ⊢ ( ( ( 𝜑 ∨ 𝜓 ) ∧ 𝜒 ) ↔ ( 𝜒 ∧ ( 𝜑 ∨ 𝜓 ) ) )
3 ancom ⊢ ( ( 𝜑 ∧ 𝜒 ) ↔ ( 𝜒 ∧ 𝜑 ) )
4 ancom ⊢ ( ( 𝜓 ∧ 𝜒 ) ↔ ( 𝜒 ∧ 𝜓 ) )
5 3 4 orbi12i ⊢ ( ( ( 𝜑 ∧ 𝜒 ) ∨ ( 𝜓 ∧ 𝜒 ) ) ↔ ( ( 𝜒 ∧ 𝜑 ) ∨ ( 𝜒 ∧ 𝜓 ) ) )
6 1 2 5 3bitr4i ⊢ ( ( ( 𝜑 ∨ 𝜓 ) ∧ 𝜒 ) ↔ ( ( 𝜑 ∧ 𝜒 ) ∨ ( 𝜓 ∧ 𝜒 ) ) )