Sr Examen

Expresión xy\/xyz\/xzp

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

    Ha introducido [src]
    (x∧y)|(x∧y∧z)|(p∧x∧z)
    $$\left(x \wedge y\right) | \left(x \wedge y \wedge z\right) | \left(p \wedge x \wedge z\right)$$

    Вы использовали:
    | - Не-и (штрих Шеффера).
    Возможно вы имели ввиду символ - Дизъюнкция (ИЛИ)?
    Посмотреть с символом ∨
    Solución detallada
    $$\left(x \wedge y\right) | \left(x \wedge y \wedge z\right) | \left(p \wedge x \wedge z\right) = \neg p \vee \neg x \vee \neg y \vee \neg z$$
    Simplificación [src]
    $$\neg p \vee \neg x \vee \neg y \vee \neg z$$
    (¬p)∨(¬x)∨(¬y)∨(¬z)
    Tabla de verdad
    +---+---+---+---+--------+
    | p | x | y | z | result |
    +===+===+===+===+========+
    | 0 | 0 | 0 | 0 | 1      |
    +---+---+---+---+--------+
    | 0 | 0 | 0 | 1 | 1      |
    +---+---+---+---+--------+
    | 0 | 0 | 1 | 0 | 1      |
    +---+---+---+---+--------+
    | 0 | 0 | 1 | 1 | 1      |
    +---+---+---+---+--------+
    | 0 | 1 | 0 | 0 | 1      |
    +---+---+---+---+--------+
    | 0 | 1 | 0 | 1 | 1      |
    +---+---+---+---+--------+
    | 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 | 1      |
    +---+---+---+---+--------+
    | 1 | 1 | 0 | 1 | 1      |
    +---+---+---+---+--------+
    | 1 | 1 | 1 | 0 | 1      |
    +---+---+---+---+--------+
    | 1 | 1 | 1 | 1 | 0      |
    +---+---+---+---+--------+
    FND [src]
    Ya está reducido a FND
    $$\neg p \vee \neg x \vee \neg y \vee \neg z$$
    (¬p)∨(¬x)∨(¬y)∨(¬z)
    FNC [src]
    Ya está reducido a FNC
    $$\neg p \vee \neg x \vee \neg y \vee \neg z$$
    (¬p)∨(¬x)∨(¬y)∨(¬z)
    FNDP [src]
    $$\neg p \vee \neg x \vee \neg y \vee \neg z$$
    (¬p)∨(¬x)∨(¬y)∨(¬z)
    FNCD [src]
    $$\neg p \vee \neg x \vee \neg y \vee \neg z$$
    (¬p)∨(¬x)∨(¬y)∨(¬z)