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

Expresión ¬(¬(¬a∨¬b∨c)∨¬a¬c∨a¬b)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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