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Theorem anddi 833
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
StepHypRef Expression
1 andir 831 . 2 (((𝜑 ∨ 𝜓) ∧ (𝜒 ∨ 𝜃)) ↔ ((𝜑 ∧ (𝜒 ∨ 𝜃)) ∨ (𝜓 ∧ (𝜒 ∨ 𝜃))))
2 andi 830 . . 3 ((𝜑 ∧ (𝜒 ∨ 𝜃)) ↔ ((𝜑 ∧ 𝜒) ∨ (𝜑 ∧ 𝜃)))
3 andi 830 . . 3 ((𝜓 ∧ (𝜒 ∨ 𝜃)) ↔ ((𝜓 ∧ 𝜒) ∨ (𝜓 ∧ 𝜃)))
42, 3orbi12i 776 . 2 (((𝜑 ∧ (𝜒 ∨ 𝜃)) ∨ (𝜓 ∧ (𝜒 ∨ 𝜃))) ↔ (((𝜑 ∧ 𝜒) ∨ (𝜑 ∧ 𝜃)) ∨ ((𝜓 ∧ 𝜒) ∨ (𝜓 ∧ 𝜃))))
51, 4bitri 184 1 (((𝜑 ∨ 𝜓) ∧ (𝜒 ∨ 𝜃)) ↔ (((𝜑 ∧ 𝜒) ∨ (𝜑 ∧ 𝜃)) ∨ ((𝜓 ∧ 𝜒) ∨ (𝜓 ∧ 𝜃))))
Colors of variables:    wff set class
This proof depends on syntax axioms:   ∧ wa 104   ↔ wb 105   ∨ wo 720
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721
This proof depends on definitions:  df-bi 117
This theorem is used by:  funun  5422  acexmidlemcase  6080  nnm00  6803
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