Expresión acvb¬cvab
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Solución
Solución detallada
(a∧b)∨(a∧c)∨(b∧¬c)=(a∧c)∨(b∧¬c)
(a∧c)∨(b∧¬c)
Tabla de verdad
+---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 0 |
+---+---+---+--------+
| 0 | 0 | 1 | 0 |
+---+---+---+--------+
| 0 | 1 | 0 | 1 |
+---+---+---+--------+
| 0 | 1 | 1 | 0 |
+---+---+---+--------+
| 1 | 0 | 0 | 0 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
(a∧c)∨(b∧¬c)
(a∨b)∧(a∨¬c)∧(b∨c)∧(c∨¬c)
(a∨b)∧(b∨c)∧(a∨(¬c))∧(c∨(¬c))
(a∨¬c)∧(b∨c)
Ya está reducido a FND
(a∧c)∨(b∧¬c)