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Lógica matemática/ ab(¬c)∨a(¬b)c∨a(¬b)(¬c)∨(¬a)(¬b)c∨(¬a)(¬b)(¬c)

Expresión ab(¬c)∨a(¬b)c∨a(¬b)(¬c)∨(¬a)(¬b)c∨(¬a)(¬b)(¬c)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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