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Lógica matemática/ xy∨(¬x⇒¬x∧¬y⇒z)

Expresión xy∨(¬x⇒¬x∧¬y⇒z)

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    Solución

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