Expresión not(a&(bornotc)ornota&b)
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
$$\left(a \wedge \left(b \vee \neg c\right)\right) \vee \left(b \wedge \neg a\right) = b \vee \left(a \wedge \neg c\right)$$
$$\neg \left(\left(a \wedge \left(b \vee \neg c\right)\right) \vee \left(b \wedge \neg a\right)\right) = \neg b \wedge \left(c \vee \neg a\right)$$
$$\neg b \wedge \left(c \vee \neg a\right)$$
Tabla de verdad
+---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 1 |
+---+---+---+--------+
| 0 | 0 | 1 | 1 |
+---+---+---+--------+
| 0 | 1 | 0 | 0 |
+---+---+---+--------+
| 0 | 1 | 1 | 0 |
+---+---+---+--------+
| 1 | 0 | 0 | 0 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 0 |
+---+---+---+--------+
| 1 | 1 | 1 | 0 |
+---+---+---+--------+
Ya está reducido a FNC
$$\neg b \wedge \left(c \vee \neg a\right)$$
$$\left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg b\right)$$
$$\neg b \wedge \left(c \vee \neg a\right)$$
$$\left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg b\right)$$