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Lógica matemática/ avb⇔¬b&c

Expresión avb⇔¬b&c

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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