Sr Examen
Expresión ¬(avbvc)v¬(¬av¬(b∧¬c))
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
(¬(a∨b∨c))∨(¬((¬a)∨(¬(b∧(¬c)))))
$$\neg \left(\neg a \vee \neg \left(b \wedge \neg c\right)\right) \vee \neg \left(a \vee b \vee c\right)$$
Solución detallada
$$\neg \left(a \vee b \vee c\right) = \neg a \wedge \neg b \wedge \neg c$$
$$\neg \left(b \wedge \neg c\right) = c \vee \neg b$$
$$\neg a \vee \neg \left(b \wedge \neg c\right) = c \vee \neg a \vee \neg b$$
$$\neg \left(\neg a \vee \neg \left(b \wedge \neg c\right)\right) = a \wedge b \wedge \neg c$$
$$\neg \left(\neg a \vee \neg \left(b \wedge \neg c\right)\right) \vee \neg \left(a \vee b \vee c\right) = \neg c \wedge \left(a \vee \neg b\right) \wedge \left(b \vee \neg a\right)$$
$$\neg \left(b \wedge \neg c\right) = c \vee \neg b$$
$$\neg a \vee \neg \left(b \wedge \neg c\right) = c \vee \neg a \vee \neg b$$
$$\neg \left(\neg a \vee \neg \left(b \wedge \neg c\right)\right) = a \wedge b \wedge \neg c$$
$$\neg \left(\neg a \vee \neg \left(b \wedge \neg c\right)\right) \vee \neg \left(a \vee b \vee c\right) = \neg c \wedge \left(a \vee \neg b\right) \wedge \left(b \vee \neg a\right)$$
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))
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))
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))