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Theorem 3ioran 995
Description: Negated triple disjunction as triple conjunction. (Contributed by Scott Fenton, 19-Apr-2011.)
Assertion
Ref Expression
3ioran (¬ (𝜑𝜓𝜒) ↔ (¬ 𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜒))

Proof of Theorem 3ioran
StepHypRef Expression
1 ioran 753 . . 3 (¬ (𝜑𝜓) ↔ (¬ 𝜑 ∧ ¬ 𝜓))
21anbi1i 458 . 2 ((¬ (𝜑𝜓) ∧ ¬ 𝜒) ↔ ((¬ 𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜒))
3 ioran 753 . . 3 (¬ ((𝜑𝜓) ∨ 𝜒) ↔ (¬ (𝜑𝜓) ∧ ¬ 𝜒))
4 df-3or 981 . . 3 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∨ 𝜒))
53, 4xchnxbir 682 . 2 (¬ (𝜑𝜓𝜒) ↔ (¬ (𝜑𝜓) ∧ ¬ 𝜒))
6 df-3an 982 . 2 ((¬ 𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜒) ↔ ((¬ 𝜑 ∧ ¬ 𝜓) ∧ ¬ 𝜒))
72, 5, 63bitr4i 212 1 (¬ (𝜑𝜓𝜒) ↔ (¬ 𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜒))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wa 104  wb 105  wo 709  w3o 979  w3a 980
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710
This theorem depends on definitions:  df-bi 117  df-3or 981  df-3an 982
This theorem is referenced by:  ne3anior  2448  onntri35  7265
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