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Lógica matemática/ ¬((p∨q)∧¬((p∧¬q)∨r))

Expresión ¬((p∨q)∧¬((p∧¬q)∨r))

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    Solución

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