Sr Examen
Expresión {¬qv[pv¬(q^p)]}^¬[¬(qvr)v(¬pvq)]
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
(p∨(¬q)∨(¬(p∧q)))∧(¬(q∨(¬p)∨(¬(q∨r))))
$$\neg \left(q \vee \neg p \vee \neg \left(q \vee r\right)\right) \wedge \left(p \vee \neg q \vee \neg \left(p \wedge q\right)\right)$$
Solución detallada
$$\neg \left(p \wedge q\right) = \neg p \vee \neg q$$
$$p \vee \neg q \vee \neg \left(p \wedge q\right) = 1$$
$$\neg \left(q \vee r\right) = \neg q \wedge \neg r$$
$$q \vee \neg p \vee \neg \left(q \vee r\right) = q \vee \neg p \vee \neg r$$
$$\neg \left(q \vee \neg p \vee \neg \left(q \vee r\right)\right) = p \wedge r \wedge \neg q$$
$$\neg \left(q \vee \neg p \vee \neg \left(q \vee r\right)\right) \wedge \left(p \vee \neg q \vee \neg \left(p \wedge q\right)\right) = p \wedge r \wedge \neg q$$
$$p \vee \neg q \vee \neg \left(p \wedge q\right) = 1$$
$$\neg \left(q \vee r\right) = \neg q \wedge \neg r$$
$$q \vee \neg p \vee \neg \left(q \vee r\right) = q \vee \neg p \vee \neg r$$
$$\neg \left(q \vee \neg p \vee \neg \left(q \vee r\right)\right) = p \wedge r \wedge \neg q$$
$$\neg \left(q \vee \neg p \vee \neg \left(q \vee r\right)\right) \wedge \left(p \vee \neg q \vee \neg \left(p \wedge q\right)\right) = p \wedge r \wedge \neg q$$
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 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+