Expresión не((p<—>неq)vr)^q
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
$$p ⇔ \neg q = \left(p \wedge \neg q\right) \vee \left(q \wedge \neg p\right)$$
$$r \vee \left(p ⇔ \neg q\right) = r \vee \left(p \wedge \neg q\right) \vee \left(q \wedge \neg p\right)$$
$$\neg \left(r \vee \left(p ⇔ \neg q\right)\right) = \neg r \wedge \left(p \vee \neg q\right) \wedge \left(q \vee \neg p\right)$$
$$q \wedge \neg \left(r \vee \left(p ⇔ \neg q\right)\right) = p \wedge q \wedge \neg r$$
$$p \wedge q \wedge \neg r$$
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 | 0 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 0 |
+---+---+---+--------+
$$p \wedge q \wedge \neg r$$
$$p \wedge q \wedge \neg r$$
Ya está reducido a FND
$$p \wedge q \wedge \neg r$$
Ya está reducido a FNC
$$p \wedge q \wedge \neg r$$