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Lógica matemática/ A+BC+(!B!C)

Expresión A+BC+(!B!C)

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    Solución

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