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Theorem andi3or 39102
Description: Distribute over triple disjunction. (Contributed by RP, 5-Jul-2021.)
Assertion
Ref Expression
andi3or ((𝜑 ∧ (𝜓𝜒𝜃)) ↔ ((𝜑𝜓) ∨ (𝜑𝜒) ∨ (𝜑𝜃)))

Proof of Theorem andi3or
StepHypRef Expression
1 andi 1031 . . 3 ((𝜑 ∧ ((𝜓𝜒) ∨ 𝜃)) ↔ ((𝜑 ∧ (𝜓𝜒)) ∨ (𝜑𝜃)))
2 andi 1031 . . . 4 ((𝜑 ∧ (𝜓𝜒)) ↔ ((𝜑𝜓) ∨ (𝜑𝜒)))
32orbi1i 938 . . 3 (((𝜑 ∧ (𝜓𝜒)) ∨ (𝜑𝜃)) ↔ (((𝜑𝜓) ∨ (𝜑𝜒)) ∨ (𝜑𝜃)))
41, 3bitri 267 . 2 ((𝜑 ∧ ((𝜓𝜒) ∨ 𝜃)) ↔ (((𝜑𝜓) ∨ (𝜑𝜒)) ∨ (𝜑𝜃)))
5 df-3or 1109 . . 3 ((𝜓𝜒𝜃) ↔ ((𝜓𝜒) ∨ 𝜃))
65anbi2i 617 . 2 ((𝜑 ∧ (𝜓𝜒𝜃)) ↔ (𝜑 ∧ ((𝜓𝜒) ∨ 𝜃)))
7 df-3or 1109 . 2 (((𝜑𝜓) ∨ (𝜑𝜒) ∨ (𝜑𝜃)) ↔ (((𝜑𝜓) ∨ (𝜑𝜒)) ∨ (𝜑𝜃)))
84, 6, 73bitr4i 295 1 ((𝜑 ∧ (𝜓𝜒𝜃)) ↔ ((𝜑𝜓) ∨ (𝜑𝜒) ∨ (𝜑𝜃)))
Colors of variables: wff setvar class
Syntax hints:  wb 198  wa 385  wo 874  w3o 1107
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 199  df-an 386  df-or 875  df-3or 1109
This theorem is referenced by:  uneqsn  39103
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