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Expresión ¬a⊕¬a⇒(¬b+c)⊕c

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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