Expresión ab⊕bc
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Solución
Solución detallada
(a∧b)⊕(b∧c)=b∧(a∨c)∧(¬a∨¬c)
b∧(a∨c)∧(¬a∨¬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 | 0 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 0 |
+---+---+---+--------+
b∧(a∨c)∧(¬a∨¬c)
(a∧b∧¬a)∨(a∧b∧¬c)∨(b∧c∧¬a)∨(b∧c∧¬c)
(a∧b∧(¬a))∨(a∧b∧(¬c))∨(b∧c∧(¬a))∨(b∧c∧(¬c))
Ya está reducido a FNC
b∧(a∨c)∧(¬a∨¬c)
(a∧b∧¬c)∨(b∧c∧¬a)