Sr Examen

Expresión (x∧y)∨(x∧y∧z)∨(p∧x∧z)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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