Sr Examen
Expresión ((p^q)v~(rvp)↔((r^p)→(qvr))^~p)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
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((p∧q)∨(¬(p∨r)))⇔((¬p)∧((p∧r)⇒(q∨r)))
$$\left(\left(\left(p \wedge r\right) \Rightarrow \left(q \vee r\right)\right) \wedge \neg p\right) ⇔ \left(\left(p \wedge q\right) \vee \neg \left(p \vee r\right)\right)$$
Solución detallada
$$\neg \left(p \vee r\right) = \neg p \wedge \neg r$$
$$\left(p \wedge q\right) \vee \neg \left(p \vee r\right) = \left(p \wedge q\right) \vee \left(\neg p \wedge \neg r\right)$$
$$\left(p \wedge r\right) \Rightarrow \left(q \vee r\right) = 1$$
$$\left(\left(p \wedge r\right) \Rightarrow \left(q \vee r\right)\right) \wedge \neg p = \neg p$$
$$\left(\left(\left(p \wedge r\right) \Rightarrow \left(q \vee r\right)\right) \wedge \neg p\right) ⇔ \left(\left(p \wedge q\right) \vee \neg \left(p \vee r\right)\right) = \left(p \wedge \neg q\right) \vee \left(\neg p \wedge \neg r\right)$$
$$\left(p \wedge q\right) \vee \neg \left(p \vee r\right) = \left(p \wedge q\right) \vee \left(\neg p \wedge \neg r\right)$$
$$\left(p \wedge r\right) \Rightarrow \left(q \vee r\right) = 1$$
$$\left(\left(p \wedge r\right) \Rightarrow \left(q \vee r\right)\right) \wedge \neg p = \neg p$$
$$\left(\left(\left(p \wedge r\right) \Rightarrow \left(q \vee r\right)\right) \wedge \neg p\right) ⇔ \left(\left(p \wedge q\right) \vee \neg \left(p \vee r\right)\right) = \left(p \wedge \neg q\right) \vee \left(\neg p \wedge \neg r\right)$$
Simplificación
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$$\left(p \wedge \neg q\right) \vee \left(\neg p \wedge \neg r\right)$$
(p∧(¬q))∨((¬p)∧(¬r))
Tabla de verdad
+---+---+---+--------+ | p | q | r | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+
FNDP
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$$\left(p \wedge \neg q\right) \vee \left(\neg p \wedge \neg r\right)$$
(p∧(¬q))∨((¬p)∧(¬r))
FNCD
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$$\left(p \vee \neg r\right) \wedge \left(\neg p \vee \neg q\right)$$
(p∨(¬r))∧((¬p)∨(¬q))
FNC
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$$\left(p \vee \neg p\right) \wedge \left(p \vee \neg r\right) \wedge \left(\neg p \vee \neg q\right) \wedge \left(\neg q \vee \neg r\right)$$
(p∨(¬p))∧(p∨(¬r))∧((¬p)∨(¬q))∧((¬q)∨(¬r))
FND
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Ya está reducido a FND
$$\left(p \wedge \neg q\right) \vee \left(\neg p \wedge \neg r\right)$$
(p∧(¬q))∨((¬p)∧(¬r))