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