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Theorem 3jaob 1234
Description: Disjunction of 3 antecedents. (Contributed by NM, 13-Sep-2011.)
Assertion
Ref Expression
3jaob (((𝜑𝜒𝜃) → 𝜓) ↔ ((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)))

Proof of Theorem 3jaob
StepHypRef Expression
1 3mix1 1108 . . . 4 (𝜑 → (𝜑𝜒𝜃))
21imim1i 59 . . 3 (((𝜑𝜒𝜃) → 𝜓) → (𝜑𝜓))
3 3mix2 1109 . . . 4 (𝜒 → (𝜑𝜒𝜃))
43imim1i 59 . . 3 (((𝜑𝜒𝜃) → 𝜓) → (𝜒𝜓))
5 3mix3 1110 . . . 4 (𝜃 → (𝜑𝜒𝜃))
65imim1i 59 . . 3 (((𝜑𝜒𝜃) → 𝜓) → (𝜃𝜓))
72, 4, 63jca 1119 . 2 (((𝜑𝜒𝜃) → 𝜓) → ((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)))
8 3jao 1233 . 2 (((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)) → ((𝜑𝜒𝜃) → 𝜓))
97, 8impbii 124 1 (((𝜑𝜒𝜃) → 𝜓) ↔ ((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 103  w3o 919  w3a 920
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 663
This theorem depends on definitions:  df-bi 115  df-3or 921  df-3an 922
This theorem is referenced by: (None)
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