Sr Examen

Expresión xyvy¬zvxzv¬yz

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

    Ha introducido [src]
    (x∧y)∨(x∧z)∨(y∧(¬z))∨(z∧(¬y))
    $$\left(x \wedge y\right) \vee \left(x \wedge z\right) \vee \left(y \wedge \neg z\right) \vee \left(z \wedge \neg y\right)$$
    Solución detallada
    $$\left(x \wedge y\right) \vee \left(x \wedge z\right) \vee \left(y \wedge \neg z\right) \vee \left(z \wedge \neg y\right) = \left(x \wedge y\right) \vee \left(y \wedge \neg z\right) \vee \left(z \wedge \neg y\right)$$
    Simplificación [src]
    $$\left(x \wedge y\right) \vee \left(y \wedge \neg z\right) \vee \left(z \wedge \neg y\right)$$
    (x∧y)∨(y∧(¬z))∨(z∧(¬y))
    Tabla de verdad
    +---+---+---+--------+
    | x | y | z | result |
    +===+===+===+========+
    | 0 | 0 | 0 | 0      |
    +---+---+---+--------+
    | 0 | 0 | 1 | 1      |
    +---+---+---+--------+
    | 0 | 1 | 0 | 1      |
    +---+---+---+--------+
    | 0 | 1 | 1 | 0      |
    +---+---+---+--------+
    | 1 | 0 | 0 | 0      |
    +---+---+---+--------+
    | 1 | 0 | 1 | 1      |
    +---+---+---+--------+
    | 1 | 1 | 0 | 1      |
    +---+---+---+--------+
    | 1 | 1 | 1 | 1      |
    +---+---+---+--------+
    FNDP [src]
    $$\left(x \wedge y\right) \vee \left(y \wedge \neg z\right) \vee \left(z \wedge \neg y\right)$$
    (x∧y)∨(y∧(¬z))∨(z∧(¬y))
    FNCD [src]
    $$\left(y \vee z\right) \wedge \left(x \vee \neg y \vee \neg z\right)$$
    (y∨z)∧(x∨(¬y)∨(¬z))
    FNC [src]
    $$\left(y \vee z\right) \wedge \left(y \vee \neg y\right) \wedge \left(x \vee y \vee z\right) \wedge \left(x \vee y \vee \neg y\right) \wedge \left(x \vee z \vee \neg z\right) \wedge \left(x \vee \neg y \vee \neg z\right) \wedge \left(y \vee z \vee \neg z\right) \wedge \left(y \vee \neg y \vee \neg z\right)$$
    (y∨z)∧(y∨(¬y))∧(x∨y∨z)∧(x∨y∨(¬y))∧(x∨z∨(¬z))∧(y∨z∨(¬z))∧(x∨(¬y)∨(¬z))∧(y∨(¬y)∨(¬z))
    FND [src]
    Ya está reducido a FND
    $$\left(x \wedge y\right) \vee \left(y \wedge \neg z\right) \vee \left(z \wedge \neg y\right)$$
    (x∧y)∨(y∧(¬z))∨(z∧(¬y))