Expresión CA+C¬BC+BA+BC¬B
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Solución
Solución detallada
b∧c∧¬b=False(a∧b)∨(a∧c)∨(c∧¬b)∨(b∧c∧¬b)=(a∧b)∨(c∧¬b)
(a∧b)∨(c∧¬b)
Tabla de verdad
+---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 0 |
+---+---+---+--------+
| 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 | 1 |
+---+---+---+--------+
(a∨c)∧(a∨¬b)∧(b∨c)∧(b∨¬b)
(a∨c)∧(b∨c)∧(a∨(¬b))∧(b∨(¬b))
(a∧b)∨(c∧¬b)
Ya está reducido a FND
(a∧b)∨(c∧¬b)
(a∨¬b)∧(b∨c)