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Theorem 3jaob 1425
Description: Disjunction of three antecedents. (Contributed by NM, 13-Sep-2011.) (Proof shortened by Hongxiu Chen, 29-Jun-2025.)
Assertion
Ref Expression
3jaob (((𝜑𝜒𝜃) → 𝜓) ↔ ((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)))

Proof of Theorem 3jaob
StepHypRef Expression
1 pm5.53 1006 . 2 ((((𝜑𝜒) ∨ 𝜃) → 𝜓) ↔ (((𝜑𝜓) ∧ (𝜒𝜓)) ∧ (𝜃𝜓)))
2 df-3or 1087 . . 3 ((𝜑𝜒𝜃) ↔ ((𝜑𝜒) ∨ 𝜃))
32imbi1i 349 . 2 (((𝜑𝜒𝜃) → 𝜓) ↔ (((𝜑𝜒) ∨ 𝜃) → 𝜓))
4 df-3an 1088 . 2 (((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)) ↔ (((𝜑𝜓) ∧ (𝜒𝜓)) ∧ (𝜃𝜓)))
51, 3, 43bitr4i 303 1 (((𝜑𝜒𝜃) → 𝜓) ↔ ((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395  wo 847  w3o 1085  w3a 1086
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 207  df-an 396  df-or 848  df-3or 1087  df-3an 1088
This theorem is referenced by: (None)
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