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