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Lógica matemática/ ¬(¬A&BvA&(Bv¬C))⇔¬B&(¬AvC)

Expresión ¬(¬A&BvA&(Bv¬C))⇔¬B&(¬AvC)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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