Metamath Proof Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >  3jao Structured version   Visualization version   GIF version

Theorem 3jao 1422
 Description: Disjunction of three antecedents. (Contributed by NM, 8-Apr-1994.)
Assertion
Ref Expression
3jao (((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)) → ((𝜑𝜒𝜃) → 𝜓))

Proof of Theorem 3jao
StepHypRef Expression
1 jao 958 . . 3 ((𝜑𝜓) → ((𝜒𝜓) → ((𝜑𝜒) → 𝜓)))
2 df-3or 1085 . . . 4 ((𝜑𝜒𝜃) ↔ ((𝜑𝜒) ∨ 𝜃))
3 jao 958 . . . 4 (((𝜑𝜒) → 𝜓) → ((𝜃𝜓) → (((𝜑𝜒) ∨ 𝜃) → 𝜓)))
42, 3syl7bi 258 . . 3 (((𝜑𝜒) → 𝜓) → ((𝜃𝜓) → ((𝜑𝜒𝜃) → 𝜓)))
51, 4syl6 35 . 2 ((𝜑𝜓) → ((𝜒𝜓) → ((𝜃𝜓) → ((𝜑𝜒𝜃) → 𝜓))))
653imp 1108 1 (((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)) → ((𝜑𝜒𝜃) → 𝜓))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∨ wo 844   ∨ 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:  3jaob  1423  3jaoi  1424  3jaod  1425
 Copyright terms: Public domain W3C validator