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