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Lógica matemática/ x(-z)by(-z)v(-x)vy

Expresión x(-z)by(-z)v(-x)vy

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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