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