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Lógica matemática/ xy->zu

Expresión xy->zu

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⌨

    Solución

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