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Lógica matemática/ ((¬x∨z)⇒y)(¬x⇒z)

Expresión ((¬x∨z)⇒y)(¬x⇒z)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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