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