Sr Examen
Lógica matemática/ xy=>zm

Expresión xy=>zm

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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