Expresión notc+a->a*notb*c+nota*b+b*c
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
(b∧c)∨(b∧¬a)∨(a∧c∧¬b)=(a∧c)∨(b∧¬a)(a∨¬c)⇒((b∧c)∨(b∧¬a)∨(a∧c∧¬b))=c∨(b∧¬a)
c∨(b∧¬a)
Tabla de verdad
+---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 0 |
+---+---+---+--------+
| 0 | 0 | 1 | 1 |
+---+---+---+--------+
| 0 | 1 | 0 | 1 |
+---+---+---+--------+
| 0 | 1 | 1 | 1 |
+---+---+---+--------+
| 1 | 0 | 0 | 0 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 0 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
c∨(b∧¬a)
(b∨c)∧(c∨¬a)
Ya está reducido a FND
c∨(b∧¬a)
(b∨c)∧(c∨¬a)