Theorem anddi 1011

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

Assertion

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

Proof of Theorem anddi

Step	Hyp	Ref	Expression
1		andir 1009	. 2 ⊢ (((𝜑 ∨ 𝜓) ∧ (𝜒 ∨ 𝜃)) ↔ ((𝜑 ∧ (𝜒 ∨ 𝜃)) ∨ (𝜓 ∧ (𝜒 ∨ 𝜃))))
2		andi 1008	. . 3 ⊢ ((𝜑 ∧ (𝜒 ∨ 𝜃)) ↔ ((𝜑 ∧ 𝜒) ∨ (𝜑 ∧ 𝜃)))
3		andi 1008	. . 3 ⊢ ((𝜓 ∧ (𝜒 ∨ 𝜃)) ↔ ((𝜓 ∧ 𝜒) ∨ (𝜓 ∧ 𝜃)))
4	2, 3	orbi12i 913	. 2 ⊢ (((𝜑 ∧ (𝜒 ∨ 𝜃)) ∨ (𝜓 ∧ (𝜒 ∨ 𝜃))) ↔ (((𝜑 ∧ 𝜒) ∨ (𝜑 ∧ 𝜃)) ∨ ((𝜓 ∧ 𝜒) ∨ (𝜓 ∧ 𝜃))))
5	1, 4	bitri 275	1 ⊢ (((𝜑 ∨ 𝜓) ∧ (𝜒 ∨ 𝜃)) ↔ (((𝜑 ∧ 𝜒) ∨ (𝜑 ∧ 𝜃)) ∨ ((𝜓 ∧ 𝜒) ∨ (𝜓 ∧ 𝜃))))

Syntax hints:   ↔ wb 206   ∧ wa 395   ∨ wo 846

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 207  df-an 396  df-or 847

This theorem is referenced by:  prnebg  4880  funun  6624  addsproplem2  28021  mulsproplem9  28168  disjxpin  32610  icoreclin  37323  undif3VD  44853