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Lógica matemática/ notb×c+b×notc

Expresión notb×c+b×notc

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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