Expresión bv(a⇔cb)va¬c
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
$$a ⇔ \left(b \wedge c\right) = \left(\neg a \wedge \neg b\right) \vee \left(\neg a \wedge \neg c\right) \vee \left(a \wedge b \wedge c\right)$$
$$b \vee \left(a \wedge \neg c\right) \vee \left(a ⇔ \left(b \wedge c\right)\right) = b \vee \neg a \vee \neg c$$
$$b \vee \neg a \vee \neg c$$
Tabla de verdad
+---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 1 |
+---+---+---+--------+
| 0 | 0 | 1 | 1 |
+---+---+---+--------+
| 0 | 1 | 0 | 1 |
+---+---+---+--------+
| 0 | 1 | 1 | 1 |
+---+---+---+--------+
| 1 | 0 | 0 | 1 |
+---+---+---+--------+
| 1 | 0 | 1 | 0 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
Ya está reducido a FNC
$$b \vee \neg a \vee \neg c$$
$$b \vee \neg a \vee \neg c$$
Ya está reducido a FND
$$b \vee \neg a \vee \neg c$$
$$b \vee \neg a \vee \neg c$$