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