Sr Examen

Expresión ab¬c+¬a¬bc+abc

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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