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