Theorem anddi 1028

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 1026	. 2 ⊢ (((𝜑 ∨ 𝜓) ∧ (𝜒 ∨ 𝜃)) ↔ ((𝜑 ∧ (𝜒 ∨ 𝜃)) ∨ (𝜓 ∧ (𝜒 ∨ 𝜃))))
2		andi 1025	. . 3 ⊢ ((𝜑 ∧ (𝜒 ∨ 𝜃)) ↔ ((𝜑 ∧ 𝜒) ∨ (𝜑 ∧ 𝜃)))
3		andi 1025	. . 3 ⊢ ((𝜓 ∧ (𝜒 ∨ 𝜃)) ↔ ((𝜓 ∧ 𝜒) ∨ (𝜓 ∧ 𝜃)))
4	2, 3	orbi12i 928	. 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))))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   ∨ wo 861

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862

This theorem is used by:  prnebg  4826  funun  6589  addsproplem2  28200  mulsproplem9  28354  disjxpin  32970  icoreclin  38044  undif3VD  45631