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Mirrors > Home > NFE Home > Th. List > nannot | GIF version |
Description: Show equivalence between negation and the Nicod version. To derive nic-dfneg 1435, apply nanbi 1294. (Contributed by Jeff Hoffman, 19-Nov-2007.) |
Ref | Expression |
---|---|
nannot | ⊢ (¬ ψ ↔ (ψ ⊼ ψ)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-nan 1288 | . . 3 ⊢ ((ψ ⊼ ψ) ↔ ¬ (ψ ∧ ψ)) | |
2 | anidm 625 | . . 3 ⊢ ((ψ ∧ ψ) ↔ ψ) | |
3 | 1, 2 | xchbinx 301 | . 2 ⊢ ((ψ ⊼ ψ) ↔ ¬ ψ) |
4 | 3 | bicomi 193 | 1 ⊢ (¬ ψ ↔ (ψ ⊼ ψ)) |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 ↔ wb 176 ∧ wa 358 ⊼ wnan 1287 |
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 177 df-an 360 df-nan 1288 |
This theorem is referenced by: nanbi 1294 trunantru 1354 falnanfal 1357 nic-dfneg 1435 |
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