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