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