Sr Examen
Expresión ¬z∨¬(x∨y)=x∨y
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Solución
Ha introducido
[src]
(x∨y)⇔((¬z)∨(¬(x∨y)))
$$\left(x \vee y\right) ⇔ \left(\neg z \vee \neg \left(x \vee y\right)\right)$$
Solución detallada
$$\neg \left(x \vee y\right) = \neg x \wedge \neg y$$
$$\neg z \vee \neg \left(x \vee y\right) = \left(\neg x \wedge \neg y\right) \vee \neg z$$
$$\left(x \vee y\right) ⇔ \left(\neg z \vee \neg \left(x \vee y\right)\right) = \neg z \wedge \left(x \vee y\right)$$
$$\neg z \vee \neg \left(x \vee y\right) = \left(\neg x \wedge \neg y\right) \vee \neg z$$
$$\left(x \vee y\right) ⇔ \left(\neg z \vee \neg \left(x \vee y\right)\right) = \neg z \wedge \left(x \vee y\right)$$
Tabla de verdad
+---+---+---+--------+ | x | y | z | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+