Sr Examen
Expresión PQ+(P´Q+PQ´)(PQ)´
El profesor se sorprenderá mucho al ver tu solución correcta😉
⌨
Solución
Ha introducido
[src]
¬((p∧q)∨(p∧q∧(¬p)∧(¬(q∨(p∧q)))))
$$\neg \left(\left(p \wedge q\right) \vee \left(p \wedge q \wedge \neg p \wedge \neg \left(q \vee \left(p \wedge q\right)\right)\right)\right)$$
Solución detallada
$$q \vee \left(p \wedge q\right) = q$$
$$\neg \left(q \vee \left(p \wedge q\right)\right) = \neg q$$
$$p \wedge q \wedge \neg p \wedge \neg \left(q \vee \left(p \wedge q\right)\right) = \text{False}$$
$$\left(p \wedge q\right) \vee \left(p \wedge q \wedge \neg p \wedge \neg \left(q \vee \left(p \wedge q\right)\right)\right) = p \wedge q$$
$$\neg \left(\left(p \wedge q\right) \vee \left(p \wedge q \wedge \neg p \wedge \neg \left(q \vee \left(p \wedge q\right)\right)\right)\right) = \neg p \vee \neg q$$
$$\neg \left(q \vee \left(p \wedge q\right)\right) = \neg q$$
$$p \wedge q \wedge \neg p \wedge \neg \left(q \vee \left(p \wedge q\right)\right) = \text{False}$$
$$\left(p \wedge q\right) \vee \left(p \wedge q \wedge \neg p \wedge \neg \left(q \vee \left(p \wedge q\right)\right)\right) = p \wedge q$$
$$\neg \left(\left(p \wedge q\right) \vee \left(p \wedge q \wedge \neg p \wedge \neg \left(q \vee \left(p \wedge q\right)\right)\right)\right) = \neg p \vee \neg q$$
Tabla de verdad
+---+---+--------+ | p | q | result | +===+===+========+ | 0 | 0 | 1 | +---+---+--------+ | 0 | 1 | 1 | +---+---+--------+ | 1 | 0 | 1 | +---+---+--------+ | 1 | 1 | 0 | +---+---+--------+