Sr Examen

Expresión a&!b&!cv!(b&c)v!(a&d)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

    Ha introducido [src]
    (¬(a∧d))∨(¬(b∧c))∨(a∧(¬b)∧(¬c))
    $$\left(a \wedge \neg b \wedge \neg c\right) \vee \neg \left(a \wedge d\right) \vee \neg \left(b \wedge c\right)$$
    Solución detallada
    $$\neg \left(a \wedge d\right) = \neg a \vee \neg d$$
    $$\neg \left(b \wedge c\right) = \neg b \vee \neg c$$
    $$\left(a \wedge \neg b \wedge \neg c\right) \vee \neg \left(a \wedge d\right) \vee \neg \left(b \wedge c\right) = \neg a \vee \neg b \vee \neg c \vee \neg d$$
    Simplificación [src]
    $$\neg a \vee \neg b \vee \neg c \vee \neg d$$
    (¬a)∨(¬b)∨(¬c)∨(¬d)
    Tabla de verdad
    +---+---+---+---+--------+
    | a | b | c | d | 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
    $$\neg a \vee \neg b \vee \neg c \vee \neg d$$
    (¬a)∨(¬b)∨(¬c)∨(¬d)
    FNDP [src]
    $$\neg a \vee \neg b \vee \neg c \vee \neg d$$
    (¬a)∨(¬b)∨(¬c)∨(¬d)
    FNCD [src]
    $$\neg a \vee \neg b \vee \neg c \vee \neg d$$
    (¬a)∨(¬b)∨(¬c)∨(¬d)
    FNC [src]
    Ya está reducido a FNC
    $$\neg a \vee \neg b \vee \neg c \vee \neg d$$
    (¬a)∨(¬b)∨(¬c)∨(¬d)