Metamath Proof Explorer


Theorem ordir

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

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

Proof

Step Hyp Ref Expression
1 ordi ⊢ ( ( 𝜒 ∨ ( 𝜑 ∧ 𝜓 ) ) ↔ ( ( 𝜒 ∨ 𝜑 ) ∧ ( 𝜒 ∨ 𝜓 ) ) )
2 orcom ⊢ ( ( ( 𝜑 ∧ 𝜓 ) ∨ 𝜒 ) ↔ ( 𝜒 ∨ ( 𝜑 ∧ 𝜓 ) ) )
3 orcom ⊢ ( ( 𝜑 ∨ 𝜒 ) ↔ ( 𝜒 ∨ 𝜑 ) )
4 orcom ⊢ ( ( 𝜓 ∨ 𝜒 ) ↔ ( 𝜒 ∨ 𝜓 ) )
5 3 4 anbi12i ⊢ ( ( ( 𝜑 ∨ 𝜒 ) ∧ ( 𝜓 ∨ 𝜒 ) ) ↔ ( ( 𝜒 ∨ 𝜑 ) ∧ ( 𝜒 ∨ 𝜓 ) ) )
6 1 2 5 3bitr4i ⊢ ( ( ( 𝜑 ∧ 𝜓 ) ∨ 𝜒 ) ↔ ( ( 𝜑 ∨ 𝜒 ) ∧ ( 𝜓 ∨ 𝜒 ) ) )