Sr Examen
Expresión not(z)<->not(x->(not(y)vz))&(y->(xvnot(z)))
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
$$y \Rightarrow \left(x \vee \neg z\right) = x \vee \neg y \vee \neg z$$
$$x \Rightarrow \left(z \vee \neg y\right) = z \vee \neg x \vee \neg y$$
$$x \not\Rightarrow \left(z \vee \neg y\right) = x \wedge y \wedge \neg z$$
$$\left(y \Rightarrow \left(x \vee \neg z\right)\right) \wedge x \not\Rightarrow \left(z \vee \neg y\right) = x \wedge y \wedge \neg z$$
$$\left(\left(y \Rightarrow \left(x \vee \neg z\right)\right) \wedge x \not\Rightarrow \left(z \vee \neg y\right)\right) ⇔ \neg z = z \vee \left(x \wedge y\right)$$
$$x \Rightarrow \left(z \vee \neg y\right) = z \vee \neg x \vee \neg y$$
$$x \not\Rightarrow \left(z \vee \neg y\right) = x \wedge y \wedge \neg z$$
$$\left(y \Rightarrow \left(x \vee \neg z\right)\right) \wedge x \not\Rightarrow \left(z \vee \neg y\right) = x \wedge y \wedge \neg z$$
$$\left(\left(y \Rightarrow \left(x \vee \neg z\right)\right) \wedge x \not\Rightarrow \left(z \vee \neg y\right)\right) ⇔ \neg z = z \vee \left(x \wedge y\right)$$
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 | +---+---+---+--------+