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