Theorem orddi 1010

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

Assertion

Ref	Expression
orddi	⊢ (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ 𝜃)) ↔ (((𝜑 ∨ 𝜒) ∧ (𝜑 ∨ 𝜃)) ∧ ((𝜓 ∨ 𝜒) ∧ (𝜓 ∨ 𝜃))))

Proof of Theorem orddi

Step	Hyp	Ref	Expression
1		ordir 1007	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ 𝜃)) ↔ ((𝜑 ∨ (𝜒 ∧ 𝜃)) ∧ (𝜓 ∨ (𝜒 ∧ 𝜃))))
2		ordi 1006	. . 3 ⊢ ((𝜑 ∨ (𝜒 ∧ 𝜃)) ↔ ((𝜑 ∨ 𝜒) ∧ (𝜑 ∨ 𝜃)))
3		ordi 1006	. . 3 ⊢ ((𝜓 ∨ (𝜒 ∧ 𝜃)) ↔ ((𝜓 ∨ 𝜒) ∧ (𝜓 ∨ 𝜃)))
4	2, 3	anbi12i 630	. 2 ⊢ (((𝜑 ∨ (𝜒 ∧ 𝜃)) ∧ (𝜓 ∨ (𝜒 ∧ 𝜃))) ↔ (((𝜑 ∨ 𝜒) ∧ (𝜑 ∨ 𝜃)) ∧ ((𝜓 ∨ 𝜒) ∧ (𝜓 ∨ 𝜃))))
5	1, 4	bitri 278	1 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ 𝜃)) ↔ (((𝜑 ∨ 𝜒) ∧ (𝜑 ∨ 𝜃)) ∧ ((𝜓 ∨ 𝜒) ∧ (𝜓 ∨ 𝜃))))

Syntax hints:   ↔ wb 209   ∧ wa 399   ∨ wo 847

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848

This theorem is referenced by:  reuprg  4619  prneimg  4765  wl-cases2-dnf  35408