Sr Examen

Expresión avbv¬b&c&dv¬b&¬c&¬dv¬b&¬c&d

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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