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