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Lógica matemática/ YZ'(Z'+Z'X)+(X'Y+X'Z)

Expresión YZ'(Z'+Z'X)+(X'Y+X'Z)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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