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Schröder, Ernst: Vorlesungen über die Algebra der Logik. Bd. 2, Abt. 2. Leipzig, 1905.

Bild:
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McColl's Anwendung des Aussagenkalkuls etc.

Die Grenzentabelle:

[Tabelle]
führt unter Wegfall mancher früheren nur w4, x7, y12, y13, y14, z7, z8 als
neue Grenzen ein. Sonach:
A = x0, 1, 3, 5, (z0' + z0 x2') w0', 3', 4

Bei Geltung von x0 ist: w0', 3', 4 = w3', 4 = w3', 4 x7', also:
A = (B1 + B2) w3', 4, wo B1 = z0' x7', 0, 1, 3, 5, B2 = z0 x2', 7', 0, 1, 3, 5.

Nun ist: x0, 1, 3, 5 = y0', 2' x1 + y0, 1 x3 + z0' y1', 2 x5, wie früher S. 547. Also:
B1 = z0' (y0', 2' · y12 x7', 1 + y0, 1 · y11' x7', 3 + y1', 2 · z7 x7', 5).

Aber bei z0' ist: y0', 2', 12 = y2', 12 = y2', 12 z7, y11', 0, 1 = y11', 1 = y11', 1 z7,
y1', 2 = y1', 2 z0', also:
B1 = z0', 7 (y2', 12 x7', 1 + y11', 1 x7', 3 + y1', 2 x7', 5).

Bei z0 ist x2', 7' = y14', 13 x2' + (y13' + y14) x7', also:
B2 = z0 {y0', 2', 14', 13 · y7' x2', 1 + (y0', 2', 13' + y0', 2', 14) y12 x7', 1 + y14', 0, 1, 13 y8 x2', 3 +
+ (y13', 0, 1 + y0, 1, 14) y11' x7', 3},
wo y0', 2', 14', 7', 13 = y7', 13 z8', y0', 2', 13', 12 = y13', 12 z8', y0', 2', 12, 14 = 0,
y14', 0, 1, 8, 13 = y14', 8 · z2', y11', 13', 0, 1 = 0, y11', 0, 1, 14 = y11', 14 z2'

bei z0 ist, somit:
B2 = z8', 0 (y7', 13 x2', 1 + y13', 12 x7', 1) + z2', 0 (y14', 8 x2', 3 + y11', 14 x7', 3).

Dies gibt:
[Formel 1] .

McColl’s Anwendung des Aussagenkalkuls etc.

Die Grenzentabelle:

[Tabelle]
führt unter Wegfall mancher früheren nur w4, x7, y12, y13, y14, z7, z8 als
neue Grenzen ein. Sonach:
A = x0, 1, 3, 5, (z0' + z0 x2') w0', 3', 4

Bei Geltung von x0 ist: w0', 3', 4 = w3', 4 = w3', 4 x7', also:
A = (B1 + B2) w3', 4, wo B1 = z0' x7', 0, 1, 3, 5, B2 = z0 x2', 7', 0, 1, 3, 5.

Nun ist: x0, 1, 3, 5 = y0', 2' x1 + y0, 1 x3 + z0' y1', 2 x5, wie früher S. 547. Also:
B1 = z0' (y0', 2' · y12 x7', 1 + y0, 1 · y11' x7', 3 + y1', 2 · z7 x7', 5).

Aber bei z0' ist: y0', 2', 12 = y2', 12 = y2', 12 z7, y11', 0, 1 = y11', 1 = y11', 1 z7,
y1', 2 = y1', 2 z0', also:
B1 = z0', 7 (y2', 12 x7', 1 + y11', 1 x7', 3 + y1', 2 x7', 5).

Bei z0 ist x2', 7' = y14', 13 x2' + (y13' + y14) x7', also:
B2 = z0 {y0', 2', 14', 13 · y7' x2', 1 + (y0', 2', 13' + y0', 2', 14) y12 x7', 1 + y14', 0, 1, 13 y8 x2', 3 +
+ (y13', 0, 1 + y0, 1, 14) y11' x7', 3},
wo y0', 2', 14', 7', 13 = y7', 13 z8', y0', 2', 13', 12 = y13', 12 z8', y0', 2', 12, 14 = 0,
y14', 0, 1, 8, 13 = y14', 8 · z2', y11', 13', 0, 1 = 0, y11', 0, 1, 14 = y11', 14 z2'

bei z0 ist, somit:
B2 = z8', 0 (y7', 13 x2', 1 + y13', 12 x7', 1) + z2', 0 (y14', 8 x2', 3 + y11', 14 x7', 3).

Dies gibt:
[Formel 1] .

<TEI>
  <text>
    <body>
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        <div n="2">
          <pb facs="#f0197" n="553"/>
          <fw place="top" type="header">McColl&#x2019;s Anwendung des Aussagenkalkuls etc.</fw><lb/>
          <p>Die Grenzentabelle:<lb/><table><row><cell/></row></table> führt unter Wegfall mancher früheren nur <hi rendition="#i">w</hi><hi rendition="#sub">4</hi>, <hi rendition="#i">x</hi><hi rendition="#sub">7</hi>, <hi rendition="#i">y</hi><hi rendition="#sub">12</hi>, <hi rendition="#i">y</hi><hi rendition="#sub">13</hi>, <hi rendition="#i">y</hi><hi rendition="#sub">14</hi>, <hi rendition="#i">z</hi><hi rendition="#sub">7</hi>, <hi rendition="#i">z</hi><hi rendition="#sub">8</hi> als<lb/>
neue Grenzen ein. Sonach:<lb/><hi rendition="#c"><hi rendition="#i">A</hi> = <hi rendition="#i">x</hi><hi rendition="#sub">0, 1, 3, 5,</hi> (<hi rendition="#i">z</hi><hi rendition="#sub">0'</hi> + <hi rendition="#i">z</hi><hi rendition="#sub">0</hi> <hi rendition="#i">x</hi><hi rendition="#sub">2'</hi>) <hi rendition="#i">w</hi><hi rendition="#sub">0', 3', 4</hi></hi></p><lb/>
          <p>Bei Geltung von <hi rendition="#i">x</hi><hi rendition="#sub">0</hi> ist: <hi rendition="#i">w</hi><hi rendition="#sub">0', 3', 4</hi> = <hi rendition="#i">w</hi><hi rendition="#sub">3', 4</hi> = <hi rendition="#i">w</hi><hi rendition="#sub">3', 4</hi> <hi rendition="#i">x</hi><hi rendition="#sub">7'</hi>, also:<lb/><hi rendition="#c"><hi rendition="#i">A</hi> = (<hi rendition="#i">B</hi><hi rendition="#sub">1</hi> + <hi rendition="#i">B</hi><hi rendition="#sub">2</hi>) <hi rendition="#i">w</hi><hi rendition="#sub">3', 4</hi>, wo <hi rendition="#i">B</hi><hi rendition="#sub">1</hi> = <hi rendition="#i">z</hi><hi rendition="#sub">0'</hi> <hi rendition="#i">x</hi><hi rendition="#sub">7', 0, 1, 3, 5</hi>, <hi rendition="#i">B</hi><hi rendition="#sub">2</hi> = <hi rendition="#i">z</hi><hi rendition="#sub">0</hi> <hi rendition="#i">x</hi><hi rendition="#sub">2', 7', 0, 1, 3, 5</hi>.</hi></p><lb/>
          <p>Nun ist: <hi rendition="#i">x</hi><hi rendition="#sub">0, 1, 3, 5</hi> = <hi rendition="#i">y</hi><hi rendition="#sub">0', 2'</hi> <hi rendition="#i">x</hi><hi rendition="#sub">1</hi> + <hi rendition="#i">y</hi><hi rendition="#sub">0, 1</hi> <hi rendition="#i">x</hi><hi rendition="#sub">3</hi> + <hi rendition="#i">z</hi><hi rendition="#sub">0'</hi> <hi rendition="#i">y</hi><hi rendition="#sub">1', 2</hi> <hi rendition="#i">x</hi><hi rendition="#sub">5</hi>, wie früher S. 547. Also:<lb/><hi rendition="#c"><hi rendition="#i">B</hi><hi rendition="#sub">1</hi> = <hi rendition="#i">z</hi><hi rendition="#sub">0'</hi> (<hi rendition="#i">y</hi><hi rendition="#sub">0', 2'</hi> · <hi rendition="#i">y</hi><hi rendition="#sub">12</hi> <hi rendition="#i">x</hi><hi rendition="#sub">7', 1</hi> + <hi rendition="#i">y</hi><hi rendition="#sub">0, 1</hi> · <hi rendition="#i">y</hi><hi rendition="#sub">11'</hi> <hi rendition="#i">x</hi><hi rendition="#sub">7', 3</hi> + <hi rendition="#i">y</hi><hi rendition="#sub">1', 2</hi> · <hi rendition="#i">z</hi><hi rendition="#sub">7</hi> <hi rendition="#i">x</hi><hi rendition="#sub">7', 5</hi>).</hi></p><lb/>
          <p>Aber bei <hi rendition="#i">z</hi><hi rendition="#sub">0'</hi> ist: <hi rendition="#i">y</hi><hi rendition="#sub">0', 2', 12</hi> = <hi rendition="#i">y</hi><hi rendition="#sub">2', 12</hi> = <hi rendition="#i">y</hi><hi rendition="#sub">2', 12</hi> <hi rendition="#i">z</hi><hi rendition="#sub">7</hi>, <hi rendition="#i">y</hi><hi rendition="#sub">11', 0, 1</hi> = <hi rendition="#i">y</hi><hi rendition="#sub">11', 1</hi> = <hi rendition="#i">y</hi><hi rendition="#sub">11', 1</hi> <hi rendition="#i">z</hi><hi rendition="#sub">7</hi>,<lb/><hi rendition="#i">y</hi><hi rendition="#sub">1', 2</hi> = <hi rendition="#i">y</hi><hi rendition="#sub">1', 2</hi> <hi rendition="#i">z</hi><hi rendition="#sub">0'</hi>, also:<lb/><hi rendition="#c"><hi rendition="#i">B</hi><hi rendition="#sub">1</hi> = <hi rendition="#i">z</hi><hi rendition="#sub">0', 7</hi> (<hi rendition="#i">y</hi><hi rendition="#sub">2', 12</hi> <hi rendition="#i">x</hi><hi rendition="#sub">7', 1</hi> + <hi rendition="#i">y</hi><hi rendition="#sub">11', 1</hi> <hi rendition="#i">x</hi><hi rendition="#sub">7', 3</hi> + <hi rendition="#i">y</hi><hi rendition="#sub">1', 2</hi> <hi rendition="#i">x</hi><hi rendition="#sub">7', 5</hi>).</hi></p><lb/>
          <p>Bei <hi rendition="#i">z</hi><hi rendition="#sub">0</hi> ist <hi rendition="#i">x</hi><hi rendition="#sub">2', 7'</hi> = <hi rendition="#i">y</hi><hi rendition="#sub">14', 13</hi> <hi rendition="#i">x</hi><hi rendition="#sub">2'</hi> + (<hi rendition="#i">y</hi><hi rendition="#sub">13'</hi> + <hi rendition="#i">y</hi><hi rendition="#sub">14</hi>) <hi rendition="#i">x</hi><hi rendition="#sub">7'</hi>, also:<lb/><hi rendition="#c"><hi rendition="#i">B</hi><hi rendition="#sub">2</hi> = <hi rendition="#i">z</hi><hi rendition="#sub">0</hi> {<hi rendition="#i">y</hi><hi rendition="#sub">0', 2', 14', 13</hi> · <hi rendition="#i">y</hi><hi rendition="#sub">7'</hi> <hi rendition="#i">x</hi><hi rendition="#sub">2', 1</hi> + (<hi rendition="#i">y</hi><hi rendition="#sub">0', 2', 13'</hi> + <hi rendition="#i">y</hi><hi rendition="#sub">0', 2', 14</hi>) <hi rendition="#i">y</hi><hi rendition="#sub">12</hi> <hi rendition="#i">x</hi><hi rendition="#sub">7', 1</hi> + <hi rendition="#i">y</hi><hi rendition="#sub">14', 0, 1, 13</hi> <hi rendition="#i">y</hi><hi rendition="#sub">8</hi> <hi rendition="#i">x</hi><hi rendition="#sub">2', 3</hi> +<lb/>
+ (<hi rendition="#i">y</hi><hi rendition="#sub">13', 0, 1</hi> + <hi rendition="#i">y</hi><hi rendition="#sub">0, 1, 14</hi>) <hi rendition="#i">y</hi><hi rendition="#sub">11'</hi> <hi rendition="#i">x</hi><hi rendition="#sub">7', 3</hi>},<lb/>
wo <hi rendition="#i">y</hi><hi rendition="#sub">0', 2', 14', 7', 13</hi> = <hi rendition="#i">y</hi><hi rendition="#sub">7', 13</hi> <hi rendition="#i">z</hi><hi rendition="#sub">8'</hi>, <hi rendition="#i">y</hi><hi rendition="#sub">0', 2', 13', 12</hi> = <hi rendition="#i">y</hi><hi rendition="#sub">13', 12</hi> <hi rendition="#i">z</hi><hi rendition="#sub">8'</hi>, <hi rendition="#i">y</hi><hi rendition="#sub">0', 2', 12, 14</hi> = 0,<lb/><hi rendition="#i">y</hi><hi rendition="#sub">14', 0, 1, 8, 13</hi> = <hi rendition="#i">y</hi><hi rendition="#sub">14', 8</hi> · <hi rendition="#i">z</hi><hi rendition="#sub">2'</hi>, <hi rendition="#i">y</hi><hi rendition="#sub">11', 13', 0, 1</hi> = 0, <hi rendition="#i">y</hi><hi rendition="#sub">11', 0, 1, 14</hi> = <hi rendition="#i">y</hi><hi rendition="#sub">11', 14</hi> <hi rendition="#i">z</hi><hi rendition="#sub">2'</hi></hi><lb/>
bei <hi rendition="#i">z</hi><hi rendition="#sub">0</hi> ist, somit:<lb/><hi rendition="#c"><hi rendition="#i">B</hi><hi rendition="#sub">2</hi> = <hi rendition="#i">z</hi><hi rendition="#sub">8', 0</hi> (<hi rendition="#i">y</hi><hi rendition="#sub">7', 13</hi> <hi rendition="#i">x</hi><hi rendition="#sub">2', 1</hi> + <hi rendition="#i">y</hi><hi rendition="#sub">13', 12</hi> <hi rendition="#i">x</hi><hi rendition="#sub">7', 1</hi>) + <hi rendition="#i">z</hi><hi rendition="#sub">2', 0</hi> (<hi rendition="#i">y</hi><hi rendition="#sub">14', 8</hi> <hi rendition="#i">x</hi><hi rendition="#sub">2', 3</hi> + <hi rendition="#i">y</hi><hi rendition="#sub">11', 14</hi> <hi rendition="#i">x</hi><hi rendition="#sub">7', 3</hi>).</hi></p><lb/>
          <p>Dies gibt:<lb/><hi rendition="#c"><formula/>.</hi><lb/></p>
        </div>
      </div>
    </body>
  </text>
</TEI>
[553/0197] McColl’s Anwendung des Aussagenkalkuls etc. Die Grenzentabelle: führt unter Wegfall mancher früheren nur w4, x7, y12, y13, y14, z7, z8 als neue Grenzen ein. Sonach: A = x0, 1, 3, 5, (z0' + z0 x2') w0', 3', 4 Bei Geltung von x0 ist: w0', 3', 4 = w3', 4 = w3', 4 x7', also: A = (B1 + B2) w3', 4, wo B1 = z0' x7', 0, 1, 3, 5, B2 = z0 x2', 7', 0, 1, 3, 5. Nun ist: x0, 1, 3, 5 = y0', 2' x1 + y0, 1 x3 + z0' y1', 2 x5, wie früher S. 547. Also: B1 = z0' (y0', 2' · y12 x7', 1 + y0, 1 · y11' x7', 3 + y1', 2 · z7 x7', 5). Aber bei z0' ist: y0', 2', 12 = y2', 12 = y2', 12 z7, y11', 0, 1 = y11', 1 = y11', 1 z7, y1', 2 = y1', 2 z0', also: B1 = z0', 7 (y2', 12 x7', 1 + y11', 1 x7', 3 + y1', 2 x7', 5). Bei z0 ist x2', 7' = y14', 13 x2' + (y13' + y14) x7', also: B2 = z0 {y0', 2', 14', 13 · y7' x2', 1 + (y0', 2', 13' + y0', 2', 14) y12 x7', 1 + y14', 0, 1, 13 y8 x2', 3 + + (y13', 0, 1 + y0, 1, 14) y11' x7', 3}, wo y0', 2', 14', 7', 13 = y7', 13 z8', y0', 2', 13', 12 = y13', 12 z8', y0', 2', 12, 14 = 0, y14', 0, 1, 8, 13 = y14', 8 · z2', y11', 13', 0, 1 = 0, y11', 0, 1, 14 = y11', 14 z2' bei z0 ist, somit: B2 = z8', 0 (y7', 13 x2', 1 + y13', 12 x7', 1) + z2', 0 (y14', 8 x2', 3 + y11', 14 x7', 3). Dies gibt: [FORMEL].

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Zitationshilfe: Schröder, Ernst: Vorlesungen über die Algebra der Logik. Bd. 2, Abt. 2. Leipzig, 1905, S. 553. In: Deutsches Textarchiv <https://www.deutschestextarchiv.de/schroeder_logik0202_1905/197>, abgerufen am 21.11.2024.