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Theorem 3jao 1196
 Description: Disjunction of 3 antecedents. (Contributed by NM, 8-Apr-1994.)
Assertion
Ref Expression
3jao (((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)) → ((𝜑𝜒𝜃) → 𝜓))

Proof of Theorem 3jao
StepHypRef Expression
1 df-3or 886 . 2 ((𝜑𝜒𝜃) ↔ ((𝜑𝜒) ∨ 𝜃))
2 jao 672 . . . 4 ((𝜑𝜓) → ((𝜒𝜓) → ((𝜑𝜒) → 𝜓)))
3 jao 672 . . . 4 (((𝜑𝜒) → 𝜓) → ((𝜃𝜓) → (((𝜑𝜒) ∨ 𝜃) → 𝜓)))
42, 3syl6 29 . . 3 ((𝜑𝜓) → ((𝜒𝜓) → ((𝜃𝜓) → (((𝜑𝜒) ∨ 𝜃) → 𝜓))))
543imp 1098 . 2 (((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)) → (((𝜑𝜒) ∨ 𝜃) → 𝜓))
61, 5syl5bi 141 1 (((𝜑𝜓) ∧ (𝜒𝜓) ∧ (𝜃𝜓)) → ((𝜑𝜒𝜃) → 𝜓))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∨ wo 629   ∨ w3o 884   ∧ w3a 885 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630 This theorem depends on definitions:  df-bi 110  df-3or 886  df-3an 887 This theorem is referenced by:  3jaob  1197  3jaoi  1198  3jaod  1199
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