Expresión notbandnotdornotaandbd
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
(¬b∧¬d)∨(b∧d∧¬a)=(b∨¬d)∧(d∨¬b)∧(¬a∨¬b)
(b∨¬d)∧(d∨¬b)∧(¬a∨¬b)
(b∨(¬d))∧(d∨(¬b))∧((¬a)∨(¬b))
Tabla de verdad
+---+---+---+--------+
| a | b | d | result |
+===+===+===+========+
| 0 | 0 | 0 | 1 |
+---+---+---+--------+
| 0 | 0 | 1 | 0 |
+---+---+---+--------+
| 0 | 1 | 0 | 0 |
+---+---+---+--------+
| 0 | 1 | 1 | 1 |
+---+---+---+--------+
| 1 | 0 | 0 | 1 |
+---+---+---+--------+
| 1 | 0 | 1 | 0 |
+---+---+---+--------+
| 1 | 1 | 0 | 0 |
+---+---+---+--------+
| 1 | 1 | 1 | 0 |
+---+---+---+--------+
(b∧¬b)∨(¬b∧¬d)∨(b∧d∧¬a)∨(b∧d∧¬b)∨(b∧¬a∧¬b)∨(d∧¬a∧¬d)∨(d∧¬b∧¬d)∨(¬a∧¬b∧¬d)
(b∧(¬b))∨((¬b)∧(¬d))∨(b∧d∧(¬a))∨(b∧d∧(¬b))∨(b∧(¬a)∧(¬b))∨(d∧(¬a)∧(¬d))∨(d∧(¬b)∧(¬d))∨((¬a)∧(¬b)∧(¬d))
Ya está reducido a FNC
(b∨¬d)∧(d∨¬b)∧(¬a∨¬b)
(b∨(¬d))∧(d∨(¬b))∧((¬a)∨(¬b))
(¬b∧¬d)∨(b∧d∧¬a)
(b∨¬d)∧(d∨¬b)∧(¬a∨¬b)
(b∨(¬d))∧(d∨(¬b))∧((¬a)∨(¬b))