Sr Examen
Expresión ~[(~p<->q)->r]
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Solución detallada
$$q ⇔ \neg p = \left(p \wedge \neg q\right) \vee \left(q \wedge \neg p\right)$$
$$\left(q ⇔ \neg p\right) \Rightarrow r = r \vee \left(p \wedge q\right) \vee \left(\neg p \wedge \neg q\right)$$
$$\left(q ⇔ \neg p\right) \not\Rightarrow r = \neg r \wedge \left(p \vee q\right) \wedge \left(\neg p \vee \neg q\right)$$
$$\left(q ⇔ \neg p\right) \Rightarrow r = r \vee \left(p \wedge q\right) \vee \left(\neg p \wedge \neg q\right)$$
$$\left(q ⇔ \neg p\right) \not\Rightarrow r = \neg r \wedge \left(p \vee q\right) \wedge \left(\neg p \vee \neg q\right)$$
Simplificación
[src]
$$\neg r \wedge \left(p \vee q\right) \wedge \left(\neg p \vee \neg q\right)$$
(¬r)∧(p∨q)∧((¬p)∨(¬q))
Tabla de verdad
+---+---+---+--------+ | p | q | r | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+
FNCD
[src]
$$\neg r \wedge \left(p \vee q\right) \wedge \left(\neg p \vee \neg q\right)$$
(¬r)∧(p∨q)∧((¬p)∨(¬q))
FNC
[src]
Ya está reducido a FNC
$$\neg r \wedge \left(p \vee q\right) \wedge \left(\neg p \vee \neg q\right)$$
(¬r)∧(p∨q)∧((¬p)∨(¬q))
FND
[src]
$$\left(p \wedge \neg p \wedge \neg r\right) \vee \left(p \wedge \neg q \wedge \neg r\right) \vee \left(q \wedge \neg p \wedge \neg r\right) \vee \left(q \wedge \neg q \wedge \neg r\right)$$
(p∧(¬p)∧(¬r))∨(p∧(¬q)∧(¬r))∨(q∧(¬p)∧(¬r))∨(q∧(¬q)∧(¬r))
FNDP
[src]
$$\left(p \wedge \neg q \wedge \neg r\right) \vee \left(q \wedge \neg p \wedge \neg r\right)$$
(p∧(¬q)∧(¬r))∨(q∧(¬p)∧(¬r))