Sr Examen

Expresión (a+b)-c=a+(b-c)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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