Sr Examen
Lógica matemática/ xy¬z∨¬xyz∨x¬y¬z

Expresión xy¬z∨¬xyz∨x¬y¬z

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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