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