Expresión CA+C+BA
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
c∨(a∧b)∨(a∧c)=c∨(a∧b)
c∨(a∧b)
Tabla de verdad
+---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 0 |
+---+---+---+--------+
| 0 | 0 | 1 | 1 |
+---+---+---+--------+
| 0 | 1 | 0 | 0 |
+---+---+---+--------+
| 0 | 1 | 1 | 1 |
+---+---+---+--------+
| 1 | 0 | 0 | 0 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
(a∨c)∧(b∨c)
(a∨c)∧(b∨c)
Ya está reducido a FND
c∨(a∧b)
c∨(a∧b)