Sr Examen

Expresión CA+C+BA

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    Solución

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