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Lógica matemática/ xyz+xayz+za

Expresión xyz+xayz+za

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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