Sr Examen

Expresión (¬z∧¬y)(z∨(¬y∧x))∨(z∨y)(¬z∧(y∨¬x))

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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