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Expresión DC¬BA+DCB¬A+DCBA

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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