Expresión notB⊕C∧A
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
¬b⊕(a∧c)=(¬a∧¬b)∨(¬b∧¬c)∨(a∧b∧c)
(¬a∧¬b)∨(¬b∧¬c)∨(a∧b∧c)
(a∧b∧c)∨((¬a)∧(¬b))∨((¬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 | 1 |
+---+---+---+--------+
| 1 | 0 | 1 | 0 |
+---+---+---+--------+
| 1 | 1 | 0 | 0 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
(a∨¬b)∧(c∨¬b)∧(b∨¬a∨¬c)
(a∨(¬b))∧(c∨(¬b))∧(b∨(¬a)∨(¬c))
(¬a∧¬b)∨(¬b∧¬c)∨(a∧b∧c)
(a∧b∧c)∨((¬a)∧(¬b))∨((¬b)∧(¬c))
Ya está reducido a FND
(¬a∧¬b)∨(¬b∧¬c)∨(a∧b∧c)
(a∧b∧c)∨((¬a)∧(¬b))∨((¬b)∧(¬c))
(a∨¬b)∧(b∨¬b)∧(c∨¬b)∧(a∨¬a∨¬b)∧(a∨¬a∨¬c)∧(a∨¬b∨¬c)∧(b∨¬a∨¬b)∧(b∨¬a∨¬c)∧(b∨¬b∨¬c)∧(c∨¬a∨¬b)∧(c∨¬a∨¬c)∧(c∨¬b∨¬c)
(a∨(¬b))∧(b∨(¬b))∧(c∨(¬b))∧(a∨(¬a)∨(¬b))∧(a∨(¬a)∨(¬c))∧(a∨(¬b)∨(¬c))∧(b∨(¬a)∨(¬b))∧(b∨(¬a)∨(¬c))∧(b∨(¬b)∨(¬c))∧(c∨(¬a)∨(¬b))∧(c∨(¬a)∨(¬c))∧(c∨(¬b)∨(¬c))