Sr Examen
Expresión ¬(¬(xy)⇔y⇔(xvz))
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
¬(y⇔(x∨z)⇔(¬(x∧y)))
$$y \not\equiv \neg \left(x \wedge y\right) \not\equiv \left(x \vee z\right)$$
Solución detallada
$$\neg \left(x \wedge y\right) = \neg x \vee \neg y$$
$$y ⇔ \neg \left(x \wedge y\right) ⇔ \left(x \vee z\right) = y \wedge z \wedge \neg x$$
$$y \not\equiv \neg \left(x \wedge y\right) \not\equiv \left(x \vee z\right) = x \vee \neg y \vee \neg z$$
$$y ⇔ \neg \left(x \wedge y\right) ⇔ \left(x \vee z\right) = y \wedge z \wedge \neg x$$
$$y \not\equiv \neg \left(x \wedge y\right) \not\equiv \left(x \vee z\right) = x \vee \neg y \vee \neg z$$
Tabla de verdad
+---+---+---+--------+ | x | y | z | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+