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