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Lógica matemática/ (¬x∨(yz))⇒((¬zw)∨y)

Expresión (¬x∨(yz))⇒((¬zw)∨y)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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