Expresión notB⊕notC*A
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
¬b⊕(a∧¬c)=(c∧¬b)∨(¬a∧¬b)∨(a∧b∧¬c)
(c∧¬b)∨(¬a∧¬b)∨(a∧b∧¬c)
(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 |
+---+---+---+--------+
(c∧¬b)∨(¬a∧¬b)∨(a∧b∧¬c)
(c∧(¬b))∨((¬a)∧(¬b))∨(a∧b∧(¬c))
Ya está reducido a FND
(c∧¬b)∨(¬a∧¬b)∨(a∧b∧¬c)
(c∧(¬b))∨((¬a)∧(¬b))∨(a∧b∧(¬c))
(a∨¬b)∧(b∨¬b)∧(¬b∨¬c)∧(a∨c∨¬a)∧(a∨c∨¬b)∧(a∨¬a∨¬b)∧(b∨c∨¬a)∧(b∨c∨¬b)∧(b∨¬a∨¬b)∧(c∨¬a∨¬c)∧(c∨¬b∨¬c)∧(¬a∨¬b∨¬c)
(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))
(a∨¬b)∧(¬b∨¬c)∧(b∨c∨¬a)
(a∨(¬b))∧((¬b)∨(¬c))∧(b∨c∨(¬a))