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| Description: Triple disjunction in terms of triple conjunction. (Contributed by NM, 8-Oct-2012.) | 
| Ref | Expression | 
|---|---|
| 3oran | ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ¬ (¬ 𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜒)) | 
| Step | Hyp | Ref | Expression | 
|---|---|---|---|
| 1 | 3ioran 1106 | . . 3 ⊢ (¬ (𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (¬ 𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜒)) | |
| 2 | 1 | con1bii 356 | . 2 ⊢ (¬ (¬ 𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜒) ↔ (𝜑 ∨ 𝜓 ∨ 𝜒)) | 
| 3 | 2 | bicomi 224 | 1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ¬ (¬ 𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜒)) | 
| Colors of variables: wff setvar class | 
| Syntax hints: ¬ wn 3 ↔ wb 206 ∨ w3o 1086 ∧ w3a 1087 | 
| 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 849 df-3or 1088 df-3an 1089 | 
| This theorem is referenced by: nolt02o 27740 nogt01o 27741 nosupbnd1lem6 27758 noinfbnd1lem6 27773 dalawlem10 39882 | 
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