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Theorem 3jaob 1423
 Description: Disjunction of three antecedents. (Contributed by NM, 13-Sep-2011.)
Assertion
Ref Expression
3jaob (((𝜑𝜒𝜃) → 𝜓) ↔ ((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)))

Proof of Theorem 3jaob
StepHypRef Expression
1 3mix1 1327 . . . 4 (𝜑 → (𝜑𝜒𝜃))
21imim1i 63 . . 3 (((𝜑𝜒𝜃) → 𝜓) → (𝜑𝜓))
3 3mix2 1328 . . . 4 (𝜒 → (𝜑𝜒𝜃))
43imim1i 63 . . 3 (((𝜑𝜒𝜃) → 𝜓) → (𝜒𝜓))
5 3mix3 1329 . . . 4 (𝜃 → (𝜑𝜒𝜃))
65imim1i 63 . . 3 (((𝜑𝜒𝜃) → 𝜓) → (𝜃𝜓))
72, 4, 63jca 1125 . 2 (((𝜑𝜒𝜃) → 𝜓) → ((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)))
8 3jao 1422 . 2 (((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)) → ((𝜑𝜒𝜃) → 𝜓))
97, 8impbii 212 1 (((𝜑𝜒𝜃) → 𝜓) ↔ ((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 209   ∨ w3o 1083   ∧ w3a 1084 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 210  df-an 400  df-or 845  df-3or 1085  df-3an 1086 This theorem is referenced by: (None)
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