Expresión (d&cv!(c⇒d&b))&bv!(d&c⇔cvb&c)⇒b
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
$$c \vee \left(b \wedge c\right) = c$$
$$\left(c \wedge d\right) ⇔ \left(c \vee \left(b \wedge c\right)\right) = d \vee \neg c$$
$$\left(c \wedge d\right) \not\equiv \left(c \vee \left(b \wedge c\right)\right) = c \wedge \neg d$$
$$c \Rightarrow \left(b \wedge d\right) = \left(b \wedge d\right) \vee \neg c$$
$$c \not\Rightarrow \left(b \wedge d\right) = c \wedge \left(\neg b \vee \neg d\right)$$
$$\left(c \wedge d\right) \vee c \not\Rightarrow \left(b \wedge d\right) = c$$
$$b \wedge \left(\left(c \wedge d\right) \vee c \not\Rightarrow \left(b \wedge d\right)\right) = b \wedge c$$
$$\left(b \wedge \left(\left(c \wedge d\right) \vee c \not\Rightarrow \left(b \wedge d\right)\right)\right) \vee \left(c \wedge d\right) \not\equiv \left(c \vee \left(b \wedge c\right)\right) = c \wedge \left(b \vee \neg d\right)$$
$$\left(\left(b \wedge \left(\left(c \wedge d\right) \vee c \not\Rightarrow \left(b \wedge d\right)\right)\right) \vee \left(c \wedge d\right) \not\equiv \left(c \vee \left(b \wedge c\right)\right)\right) \Rightarrow b = b \vee d \vee \neg c$$
Tabla de verdad
+---+---+---+--------+
| b | c | d | result |
+===+===+===+========+
| 0 | 0 | 0 | 1 |
+---+---+---+--------+
| 0 | 0 | 1 | 1 |
+---+---+---+--------+
| 0 | 1 | 0 | 0 |
+---+---+---+--------+
| 0 | 1 | 1 | 1 |
+---+---+---+--------+
| 1 | 0 | 0 | 1 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
Ya está reducido a FND
$$b \vee d \vee \neg c$$
Ya está reducido a FNC
$$b \vee d \vee \neg c$$