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| Mirrors > Home > MPE Home > Th. List > nannot | Structured version Visualization version GIF version | ||
| Description: Show equivalence between negation and the Nicod version. To derive nic-dfneg 1595, apply nanbi 1454. (Contributed by Jeff Hoffman, 19-Nov-2007.) |
| Ref | Expression |
|---|---|
| nannot | ⊢ (¬ 𝜓 ↔ (𝜓 ⊼ 𝜓)) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | df-nan 1448 | . . 3 ⊢ ((𝜓 ⊼ 𝜓) ↔ ¬ (𝜓 ∧ 𝜓)) | |
| 2 | anidm 676 | . . 3 ⊢ ((𝜓 ∧ 𝜓) ↔ 𝜓) | |
| 3 | 1, 2 | xchbinx 324 | . 2 ⊢ ((𝜓 ⊼ 𝜓) ↔ ¬ 𝜓) |
| 4 | 3 | bicomi 214 | 1 ⊢ (¬ 𝜓 ↔ (𝜓 ⊼ 𝜓)) |
| Colors of variables: wff setvar class |
| Syntax hints: ¬ wn 3 ↔ wb 196 ∧ wa 384 ⊼ wnan 1447 |
| 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 197 df-an 386 df-nan 1448 |
| This theorem is referenced by: nanbi 1454 trunantru 1524 falnanfal 1527 nic-dfneg 1595 andnand1 32398 imnand2 32399 |
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