Expresión ((a&¬b)v(c&¬a))v(¬(¬(a&¬b)vc)&a)
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
¬(a∧¬b)=b∨¬ac∨¬(a∧¬b)=b∨c∨¬a¬(c∨¬(a∧¬b))=a∧¬b∧¬ca∧¬(c∨¬(a∧¬b))=a∧¬b∧¬c(a∧¬b)∨(a∧¬(c∨¬(a∧¬b)))∨(c∧¬a)=(a∧¬b)∨(c∧¬a)
(a∧¬b)∨(c∧¬a)
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 | 1 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 0 |
+---+---+---+--------+
| 1 | 1 | 1 | 0 |
+---+---+---+--------+
(a∨c)∧(a∨¬a)∧(c∨¬b)∧(¬a∨¬b)
(a∨c)∧(a∨(¬a))∧(c∨(¬b))∧((¬a)∨(¬b))
(a∨c)∧(¬a∨¬b)
(a∧¬b)∨(c∧¬a)
Ya está reducido a FND
(a∧¬b)∨(c∧¬a)