Sr Examen
Expresión BA+CB+C~B~A
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
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a⇔b⇔(c∨(a∧b)∨(b∧c))
$$a ⇔ b ⇔ \left(c \vee \left(a \wedge b\right) \vee \left(b \wedge c\right)\right)$$
Solución detallada
$$c \vee \left(a \wedge b\right) \vee \left(b \wedge c\right) = c \vee \left(a \wedge b\right)$$
$$a ⇔ b ⇔ \left(c \vee \left(a \wedge b\right) \vee \left(b \wedge c\right)\right) = \left(a \vee \neg b\right) \wedge \left(a \vee \neg c\right) \wedge \left(b \vee \neg a\right)$$
$$a ⇔ b ⇔ \left(c \vee \left(a \wedge b\right) \vee \left(b \wedge c\right)\right) = \left(a \vee \neg b\right) \wedge \left(a \vee \neg c\right) \wedge \left(b \vee \neg a\right)$$
Simplificación
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$$\left(a \vee \neg b\right) \wedge \left(a \vee \neg c\right) \wedge \left(b \vee \neg a\right)$$
(a∨(¬b))∧(a∨(¬c))∧(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 | 1 | +---+---+---+--------+
FNDP
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$$\left(a \wedge b\right) \vee \left(\neg a \wedge \neg b \wedge \neg c\right)$$
(a∧b)∨((¬a)∧(¬b)∧(¬c))
FND
[src]
$$\left(a \wedge b\right) \vee \left(a \wedge \neg a\right) \vee \left(a \wedge b \wedge \neg b\right) \vee \left(a \wedge b \wedge \neg c\right) \vee \left(a \wedge \neg a \wedge \neg b\right) \vee \left(a \wedge \neg a \wedge \neg c\right) \vee \left(b \wedge \neg b \wedge \neg c\right) \vee \left(\neg a \wedge \neg b \wedge \neg c\right)$$
(a∧b)∨(a∧(¬a))∨(a∧b∧(¬b))∨(a∧b∧(¬c))∨(a∧(¬a)∧(¬b))∨(a∧(¬a)∧(¬c))∨(b∧(¬b)∧(¬c))∨((¬a)∧(¬b)∧(¬c))
FNCD
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$$\left(a \vee \neg b\right) \wedge \left(a \vee \neg c\right) \wedge \left(b \vee \neg a\right)$$
(a∨(¬b))∧(a∨(¬c))∧(b∨(¬a))
FNC
[src]
Ya está reducido a FNC
$$\left(a \vee \neg b\right) \wedge \left(a \vee \neg c\right) \wedge \left(b \vee \neg a\right)$$
(a∨(¬b))∧(a∨(¬c))∧(b∨(¬a))