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