Sr Examen

Expresión ¬(a∨b1)∨b2∨c

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

    Ha introducido [src]
    b2∨c∨(¬(a∨b1))
    b2c¬(ab1)b_{2} \vee c \vee \neg \left(a \vee b_{1}\right)
    Solución detallada
    ¬(ab1)=¬a¬b1\neg \left(a \vee b_{1}\right) = \neg a \wedge \neg b_{1}
    b2c¬(ab1)=b2c(¬a¬b1)b_{2} \vee c \vee \neg \left(a \vee b_{1}\right) = b_{2} \vee c \vee \left(\neg a \wedge \neg b_{1}\right)
    Simplificación [src]
    b2c(¬a¬b1)b_{2} \vee c \vee \left(\neg a \wedge \neg b_{1}\right)
    b2∨c∨((¬a)∧(¬b1))
    Tabla de verdad
    +---+----+----+---+--------+
    | a | b1 | b2 | c | 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 | 0      |
    +---+----+----+---+--------+
    | 0 | 1  | 0  | 1 | 1      |
    +---+----+----+---+--------+
    | 0 | 1  | 1  | 0 | 1      |
    +---+----+----+---+--------+
    | 0 | 1  | 1  | 1 | 1      |
    +---+----+----+---+--------+
    | 1 | 0  | 0  | 0 | 0      |
    +---+----+----+---+--------+
    | 1 | 0  | 0  | 1 | 1      |
    +---+----+----+---+--------+
    | 1 | 0  | 1  | 0 | 1      |
    +---+----+----+---+--------+
    | 1 | 0  | 1  | 1 | 1      |
    +---+----+----+---+--------+
    | 1 | 1  | 0  | 0 | 0      |
    +---+----+----+---+--------+
    | 1 | 1  | 0  | 1 | 1      |
    +---+----+----+---+--------+
    | 1 | 1  | 1  | 0 | 1      |
    +---+----+----+---+--------+
    | 1 | 1  | 1  | 1 | 1      |
    +---+----+----+---+--------+
    FNDP [src]
    b2c(¬a¬b1)b_{2} \vee c \vee \left(\neg a \wedge \neg b_{1}\right)
    b2∨c∨((¬a)∧(¬b1))
    FNCD [src]
    (b2c¬a)(b2c¬b1)\left(b_{2} \vee c \vee \neg a\right) \wedge \left(b_{2} \vee c \vee \neg b_{1}\right)
    (b2∨c∨(¬a))∧(b2∨c∨(¬b1))
    FND [src]
    Ya está reducido a FND
    b2c(¬a¬b1)b_{2} \vee c \vee \left(\neg a \wedge \neg b_{1}\right)
    b2∨c∨((¬a)∧(¬b1))
    FNC [src]
    (b2c¬a)(b2c¬b1)\left(b_{2} \vee c \vee \neg a\right) \wedge \left(b_{2} \vee c \vee \neg b_{1}\right)
    (b2∨c∨(¬a))∧(b2∨c∨(¬b1))