Sr Examen

Expresión avb(b&c)=(avb)&(avc)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

    Ha introducido [src]
    (a∨(b∧c))⇔((a∨b)∧(a∨c))
    ((ab)(ac))(a(bc))\left(\left(a \vee b\right) \wedge \left(a \vee c\right)\right) ⇔ \left(a \vee \left(b \wedge c\right)\right)
    Solución detallada
    (ab)(ac)=a(bc)\left(a \vee b\right) \wedge \left(a \vee c\right) = a \vee \left(b \wedge c\right)
    ((ab)(ac))(a(bc))=1\left(\left(a \vee b\right) \wedge \left(a \vee c\right)\right) ⇔ \left(a \vee \left(b \wedge c\right)\right) = 1
    Simplificación [src]
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    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 | 1      |
    +---+---+---+--------+
    | 1 | 1 | 0 | 1      |
    +---+---+---+--------+
    | 1 | 1 | 1 | 1      |
    +---+---+---+--------+
    FNDP [src]
    1
    1
    FNC [src]
    Ya está reducido a FNC
    1
    1
    FNCD [src]
    1
    1
    FND [src]
    Ya está reducido a FND
    1
    1