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