Theorem orddi 1006

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 1003	. 2 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ 𝜃)) ↔ ((𝜑 ∨ (𝜒 ∧ 𝜃)) ∧ (𝜓 ∨ (𝜒 ∧ 𝜃))))
2		ordi 1002	. . 3 ⊢ ((𝜑 ∨ (𝜒 ∧ 𝜃)) ↔ ((𝜑 ∨ 𝜒) ∧ (𝜑 ∨ 𝜃)))
3		ordi 1002	. . 3 ⊢ ((𝜓 ∨ (𝜒 ∧ 𝜃)) ↔ ((𝜓 ∨ 𝜒) ∧ (𝜓 ∨ 𝜃)))
4	2, 3	anbi12i 626	. 2 ⊢ (((𝜑 ∨ (𝜒 ∧ 𝜃)) ∧ (𝜓 ∨ (𝜒 ∧ 𝜃))) ↔ (((𝜑 ∨ 𝜒) ∧ (𝜑 ∨ 𝜃)) ∧ ((𝜓 ∨ 𝜒) ∧ (𝜓 ∨ 𝜃))))
5	1, 4	bitri 274	1 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜒 ∧ 𝜃)) ↔ (((𝜑 ∨ 𝜒) ∧ (𝜑 ∨ 𝜃)) ∧ ((𝜓 ∨ 𝜒) ∧ (𝜓 ∨ 𝜃))))

Syntax hints:   ↔ wb 205   ∧ wa 395   ∨ wo 843

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 206  df-an 396  df-or 844

This theorem is referenced by:  reuprg  4636  prneimg  4782  wl-cases2-dnf  35598