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