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Lógica matemática/ {¬qv[pv¬(q^p)]}^¬[¬(qvr)v(¬pvq)]

Expresión {¬qv[pv¬(q^p)]}^¬[¬(qvr)v(¬pvq)]

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

    Ha introducido [src] 
    (p∨(¬q)∨(¬(p∧q)))∧(¬(q∨(¬p)∨(¬(q∨r))))
    ¬(q∨¬p∨¬(q∨r))∧(p∨¬q∨¬(p∧q))\neg \left(q \vee \neg p \vee \neg \left(q \vee r\right)\right) \wedge \left(p \vee \neg q \vee \neg \left(p \wedge q\right)\right)
    Solución detallada
    ¬(p∧q)=¬p∨¬q\neg \left(p \wedge q\right) = \neg p \vee \neg q
    p∨¬q∨¬(p∧q)=1p \vee \neg q \vee \neg \left(p \wedge q\right) = 1
    ¬(q∨r)=¬q∧¬r\neg \left(q \vee r\right) = \neg q \wedge \neg r
    q∨¬p∨¬(q∨r)=q∨¬p∨¬rq \vee \neg p \vee \neg \left(q \vee r\right) = q \vee \neg p \vee \neg r
    ¬(q∨¬p∨¬(q∨r))=p∧r∧¬q\neg \left(q \vee \neg p \vee \neg \left(q \vee r\right)\right) = p \wedge r \wedge \neg q
    ¬(q∨¬p∨¬(q∨r))∧(p∨¬q∨¬(p∧q))=p∧r∧¬q\neg \left(q \vee \neg p \vee \neg \left(q \vee r\right)\right) \wedge \left(p \vee \neg q \vee \neg \left(p \wedge q\right)\right) = p \wedge r \wedge \neg q
    Simplificación [src]
    p∧r∧¬qp \wedge r \wedge \neg q 
    p∧r∧(¬q)
    Tabla de verdad 
    +---+---+---+--------+
| p | q | r | result |
+===+===+===+========+
| 0 | 0 | 0 | 0      |
+---+---+---+--------+
| 0 | 0 | 1 | 0      |
+---+---+---+--------+
| 0 | 1 | 0 | 0      |
+---+---+---+--------+
| 0 | 1 | 1 | 0      |
+---+---+---+--------+
| 1 | 0 | 0 | 0      |
+---+---+---+--------+
| 1 | 0 | 1 | 1      |
+---+---+---+--------+
| 1 | 1 | 0 | 0      |
+---+---+---+--------+
| 1 | 1 | 1 | 0      |
+---+---+---+--------+
    FND [src]
    Ya está reducido a FND
    p∧r∧¬qp \wedge r \wedge \neg q 
    p∧r∧(¬q)
    FNC [src]
    Ya está reducido a FNC
    p∧r∧¬qp \wedge r \wedge \neg q 
    p∧r∧(¬q)
    FNCD [src]
    p∧r∧¬qp \wedge r \wedge \neg q 
    p∧r∧(¬q)
    FNDP [src]
    p∧r∧¬qp \wedge r \wedge \neg q 
    p∧r∧(¬q)
    © Sr Examen – Калькулятор