Sr Examen

Expresión c&(!av!c)va&b&!c

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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