Expresión notB⊕notC*A
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
$$\neg b ⊕ \left(a \wedge \neg c\right) = \left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg b\right) \vee \left(a \wedge b \wedge \neg c\right)$$
$$\left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg b\right) \vee \left(a \wedge b \wedge \neg c\right)$$
(c∧(¬b))∨((¬a)∧(¬b))∨(a∧b∧(¬c))
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 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 0 |
+---+---+---+--------+
$$\left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg b\right) \vee \left(a \wedge b \wedge \neg c\right)$$
(c∧(¬b))∨((¬a)∧(¬b))∨(a∧b∧(¬c))
Ya está reducido a FND
$$\left(c \wedge \neg b\right) \vee \left(\neg a \wedge \neg b\right) \vee \left(a \wedge b \wedge \neg c\right)$$
(c∧(¬b))∨((¬a)∧(¬b))∨(a∧b∧(¬c))
$$\left(a \vee \neg b\right) \wedge \left(b \vee \neg b\right) \wedge \left(\neg b \vee \neg c\right) \wedge \left(a \vee c \vee \neg a\right) \wedge \left(a \vee c \vee \neg b\right) \wedge \left(a \vee \neg a \vee \neg b\right) \wedge \left(b \vee c \vee \neg a\right) \wedge \left(b \vee c \vee \neg b\right) \wedge \left(b \vee \neg a \vee \neg b\right) \wedge \left(c \vee \neg a \vee \neg c\right) \wedge \left(c \vee \neg b \vee \neg c\right) \wedge \left(\neg a \vee \neg b \vee \neg c\right)$$
(a∨(¬b))∧(b∨(¬b))∧((¬b)∨(¬c))∧(a∨c∨(¬a))∧(a∨c∨(¬b))∧(b∨c∨(¬a))∧(b∨c∨(¬b))∧(a∨(¬a)∨(¬b))∧(b∨(¬a)∨(¬b))∧(c∨(¬a)∨(¬c))∧(c∨(¬b)∨(¬c))∧((¬a)∨(¬b)∨(¬c))
$$\left(a \vee \neg b\right) \wedge \left(\neg b \vee \neg c\right) \wedge \left(b \vee c \vee \neg a\right)$$
(a∨(¬b))∧((¬b)∨(¬c))∧(b∨c∨(¬a))