Expresión (not(not(c)+not(a+b)))&(c+not(a+not(b)))+(not(c)+not(a+b))&(not(c+not(a+not(b))))
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
$$\neg \left(a \vee \neg b\right) = b \wedge \neg a$$
$$c \vee \neg \left(a \vee \neg b\right) = c \vee \left(b \wedge \neg a\right)$$
$$\neg \left(a \vee b\right) = \neg a \wedge \neg b$$
$$\neg c \vee \neg \left(a \vee b\right) = \left(\neg a \wedge \neg b\right) \vee \neg c$$
$$\neg \left(\neg c \vee \neg \left(a \vee b\right)\right) = c \wedge \left(a \vee b\right)$$
$$\neg \left(\neg c \vee \neg \left(a \vee b\right)\right) \wedge \left(c \vee \neg \left(a \vee \neg b\right)\right) = c \wedge \left(a \vee b\right)$$
$$\neg \left(c \vee \neg \left(a \vee \neg b\right)\right) = \neg c \wedge \left(a \vee \neg b\right)$$
$$\neg \left(c \vee \neg \left(a \vee \neg b\right)\right) \wedge \left(\neg c \vee \neg \left(a \vee b\right)\right) = \neg c \wedge \left(a \vee \neg b\right)$$
$$\left(\neg \left(\neg c \vee \neg \left(a \vee b\right)\right) \wedge \left(c \vee \neg \left(a \vee \neg b\right)\right)\right) \vee \left(\neg \left(c \vee \neg \left(a \vee \neg b\right)\right) \wedge \left(\neg c \vee \neg \left(a \vee b\right)\right)\right) = a \vee \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right)$$
$$a \vee \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right)$$
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 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
$$\left(a \vee b \vee \neg b\right) \wedge \left(a \vee b \vee \neg c\right) \wedge \left(a \vee c \vee \neg b\right) \wedge \left(a \vee c \vee \neg c\right)$$
(a∨b∨(¬b))∧(a∨b∨(¬c))∧(a∨c∨(¬b))∧(a∨c∨(¬c))
$$a \vee \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right)$$
Ya está reducido a FND
$$a \vee \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right)$$
$$\left(a \vee b \vee \neg c\right) \wedge \left(a \vee c \vee \neg b\right)$$