Sr Examen

Expresión (¬av¬bv¬c)&(¬a&¬b&c)v(¬a&b&¬c)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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