Sr Examen
Expresión ¬C⇔¬B∨A
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Solución detallada
$$\neg c ⇔ \left(a \vee \neg b\right) = \left(a \wedge \neg c\right) \vee \left(\neg b \wedge \neg c\right) \vee \left(b \wedge c \wedge \neg a\right)$$
Simplificación
[src]
$$\left(a \wedge \neg c\right) \vee \left(\neg b \wedge \neg c\right) \vee \left(b \wedge c \wedge \neg a\right)$$
(a∧(¬c))∨((¬b)∧(¬c))∨(b∧c∧(¬a))
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+
FNDP
[src]
$$\left(a \wedge \neg c\right) \vee \left(\neg b \wedge \neg c\right) \vee \left(b \wedge c \wedge \neg a\right)$$
(a∧(¬c))∨((¬b)∧(¬c))∨(b∧c∧(¬a))
FNCD
[src]
$$\left(b \vee \neg c\right) \wedge \left(\neg a \vee \neg c\right) \wedge \left(a \vee c \vee \neg b\right)$$
(b∨(¬c))∧((¬a)∨(¬c))∧(a∨c∨(¬b))
FND
[src]
Ya está reducido a FND
$$\left(a \wedge \neg c\right) \vee \left(\neg b \wedge \neg c\right) \vee \left(b \wedge c \wedge \neg a\right)$$
(a∧(¬c))∨((¬b)∧(¬c))∨(b∧c∧(¬a))
FNC
[src]
$$\left(b \vee \neg c\right) \wedge \left(c \vee \neg c\right) \wedge \left(\neg a \vee \neg c\right) \wedge \left(a \vee b \vee \neg b\right) \wedge \left(a \vee b \vee \neg c\right) \wedge \left(a \vee c \vee \neg b\right) \wedge \left(a \vee c \vee \neg c\right) \wedge \left(a \vee \neg a \vee \neg b\right) \wedge \left(a \vee \neg a \vee \neg c\right) \wedge \left(b \vee \neg b \vee \neg c\right) \wedge \left(c \vee \neg b \vee \neg c\right) \wedge \left(\neg a \vee \neg b \vee \neg c\right)$$
(b∨(¬c))∧(c∨(¬c))∧((¬a)∨(¬c))∧(a∨b∨(¬b))∧(a∨b∨(¬c))∧(a∨c∨(¬b))∧(a∨c∨(¬c))∧(a∨(¬a)∨(¬b))∧(a∨(¬a)∨(¬c))∧(b∨(¬b)∨(¬c))∧(c∨(¬b)∨(¬c))∧((¬a)∨(¬b)∨(¬c))