Sr Examen

Expresión ab^+c+(a+b)c+da

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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