Expresión ¬P∨(¬Q∨R)∧(Q∨¬R)
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
(q∨¬r)∧(r∨¬q)=(q∧r)∨(¬q∧¬r)((q∨¬r)∧(r∨¬q))∨¬p=(q∧r)∨(¬q∧¬r)∨¬p
(q∧r)∨(¬q∧¬r)∨¬p
Tabla de verdad
+---+---+---+--------+
| p | q | r | result |
+===+===+===+========+
| 0 | 0 | 0 | 1 |
+---+---+---+--------+
| 0 | 0 | 1 | 1 |
+---+---+---+--------+
| 0 | 1 | 0 | 1 |
+---+---+---+--------+
| 0 | 1 | 1 | 1 |
+---+---+---+--------+
| 1 | 0 | 0 | 1 |
+---+---+---+--------+
| 1 | 0 | 1 | 0 |
+---+---+---+--------+
| 1 | 1 | 0 | 0 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
(q∧r)∨(¬q∧¬r)∨¬p
(q∨¬p∨¬r)∧(r∨¬p∨¬q)
(q∨(¬p)∨(¬r))∧(r∨(¬p)∨(¬q))
(q∨¬p∨¬q)∧(q∨¬p∨¬r)∧(r∨¬p∨¬q)∧(r∨¬p∨¬r)
(q∨(¬p)∨(¬q))∧(q∨(¬p)∨(¬r))∧(r∨(¬p)∨(¬q))∧(r∨(¬p)∨(¬r))
Ya está reducido a FND
(q∧r)∨(¬q∧¬r)∨¬p