Sr Examen

Expresión АΛBΛĀ

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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