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Lógica matemática/ avb》(!z》y)

Expresión avb》(!z》y)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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