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Lógica matemática/ bc+(!b!c)

Expresión bc+(!b!c)

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    Solución

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