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