Sr Examen

Expresión (¬x+y+z)(u+y)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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