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