Sr Examen

Expresión notcand(a⇔b)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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