Expresión (not(A)vB)&(not(C)vnot(B))=>(not(B)vnot(A))
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
$$\left(b \vee \neg a\right) \wedge \left(\neg b \vee \neg c\right) = \left(b \wedge \neg c\right) \vee \left(\neg a \wedge \neg b\right)$$
$$\left(\left(b \vee \neg a\right) \wedge \left(\neg b \vee \neg c\right)\right) \Rightarrow \left(\neg a \vee \neg b\right) = c \vee \neg a \vee \neg b$$
$$c \vee \neg a \vee \neg b$$
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 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 0 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
$$c \vee \neg a \vee \neg b$$
Ya está reducido a FND
$$c \vee \neg a \vee \neg b$$
$$c \vee \neg a \vee \neg b$$
Ya está reducido a FNC
$$c \vee \neg a \vee \neg b$$