Sr Examen

Expresión F=¬(A&B&C)v(B&Cv¬A)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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