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Lógica matemática/ notz<->not(x->(notyvz))^(y->(xvnotz))

Expresión notz<->not(x->(notyvz))^(y->(xvnotz))

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

    Ha introducido [src] 
    (¬z)⇔((y⇒(x∨(¬z)))∧(¬(x⇒(z∨(¬y)))))
    $$\left(\left(y \Rightarrow \left(x \vee \neg z\right)\right) \wedge x \not\Rightarrow \left(z \vee \neg y\right)\right) ⇔ \neg z$$
    Solución detallada
    $$y \Rightarrow \left(x \vee \neg z\right) = x \vee \neg y \vee \neg z$$
    $$x \Rightarrow \left(z \vee \neg y\right) = z \vee \neg x \vee \neg y$$
    $$x \not\Rightarrow \left(z \vee \neg y\right) = x \wedge y \wedge \neg z$$
    $$\left(y \Rightarrow \left(x \vee \neg z\right)\right) \wedge x \not\Rightarrow \left(z \vee \neg y\right) = x \wedge y \wedge \neg z$$
    $$\left(\left(y \Rightarrow \left(x \vee \neg z\right)\right) \wedge x \not\Rightarrow \left(z \vee \neg y\right)\right) ⇔ \neg z = z \vee \left(x \wedge y\right)$$
    Simplificación [src]
    $$z \vee \left(x \wedge y\right)$$ 
    z∨(x∧y)
    Tabla de verdad 
    +---+---+---+--------+
| x | y | z | result |
+===+===+===+========+
| 0 | 0 | 0 | 0      |
+---+---+---+--------+
| 0 | 0 | 1 | 1      |
+---+---+---+--------+
| 0 | 1 | 0 | 0      |
+---+---+---+--------+
| 0 | 1 | 1 | 1      |
+---+---+---+--------+
| 1 | 0 | 0 | 0      |
+---+---+---+--------+
| 1 | 0 | 1 | 1      |
+---+---+---+--------+
| 1 | 1 | 0 | 1      |
+---+---+---+--------+
| 1 | 1 | 1 | 1      |
+---+---+---+--------+
    FND [src]
    Ya está reducido a FND
    $$z \vee \left(x \wedge y\right)$$ 
    z∨(x∧y)
    FNC [src]
    $$\left(x \vee z\right) \wedge \left(y \vee z\right)$$ 
    (x∨z)∧(y∨z)
    FNCD [src]
    $$\left(x \vee z\right) \wedge \left(y \vee z\right)$$ 
    (x∨z)∧(y∨z)
    FNDP [src]
    $$z \vee \left(x \wedge y\right)$$ 
    z∨(x∧y)
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