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Lógica matemática/ ¬(xyz+¬(y¬z)¬x)

Expresión ¬(xyz+¬(y¬z)¬x)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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