Sr Examen
Expresión ¬(¬x∨y∧z)∧(¬(x)∨¬y)∧x∨(y∧¬z)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
(y∧(¬z))∨(x∧((¬x)∨(¬y))∧(¬((¬x)∨(y∧z))))
$$\left(y \wedge \neg z\right) \vee \left(x \wedge \neg \left(\left(y \wedge z\right) \vee \neg x\right) \wedge \left(\neg x \vee \neg y\right)\right)$$
Solución detallada
$$\neg \left(\left(y \wedge z\right) \vee \neg x\right) = x \wedge \left(\neg y \vee \neg z\right)$$
$$x \wedge \neg \left(\left(y \wedge z\right) \vee \neg x\right) \wedge \left(\neg x \vee \neg y\right) = x \wedge \neg y$$
$$\left(y \wedge \neg z\right) \vee \left(x \wedge \neg \left(\left(y \wedge z\right) \vee \neg x\right) \wedge \left(\neg x \vee \neg y\right)\right) = \left(x \wedge \neg y\right) \vee \left(y \wedge \neg z\right)$$
$$x \wedge \neg \left(\left(y \wedge z\right) \vee \neg x\right) \wedge \left(\neg x \vee \neg y\right) = x \wedge \neg y$$
$$\left(y \wedge \neg z\right) \vee \left(x \wedge \neg \left(\left(y \wedge z\right) \vee \neg x\right) \wedge \left(\neg x \vee \neg y\right)\right) = \left(x \wedge \neg y\right) \vee \left(y \wedge \neg z\right)$$
Simplificación
[src]
$$\left(x \wedge \neg y\right) \vee \left(y \wedge \neg z\right)$$
(x∧(¬y))∨(y∧(¬z))
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 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+
FND
[src]
Ya está reducido a FND
$$\left(x \wedge \neg y\right) \vee \left(y \wedge \neg z\right)$$
(x∧(¬y))∨(y∧(¬z))
FNC
[src]
$$\left(x \vee y\right) \wedge \left(x \vee \neg z\right) \wedge \left(y \vee \neg y\right) \wedge \left(\neg y \vee \neg z\right)$$
(x∨y)∧(x∨(¬z))∧(y∨(¬y))∧((¬y)∨(¬z))