Expresión ¬(¬((A∨B)∧C))∨¬(¬((a∧b)∨(b∧C)))
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
¬(c∧(a∨b))=(¬a∧¬b)∨¬c¬(¬(c∧(a∨b)))=c∧(a∨b)(a∧b)∨(b∧c)=b∧(a∨c)¬((a∧b)∨(b∧c))=(¬a∧¬c)∨¬b¬(¬((a∧b)∨(b∧c)))=b∧(a∨c)¬(¬(c∧(a∨b)))∨¬(¬((a∧b)∨(b∧c)))=(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 | 0 |
+---+---+---+--------+
| 0 | 1 | 1 | 1 |
+---+---+---+--------+
| 1 | 0 | 0 | 0 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
(a∨b)∧(a∨c)∧(b∨c)
(a∧b)∨(a∧c)∨(b∧c)
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)