Sr Examen
Expresión Р˄(Q↔R)≡(Р˄Q)↔(Р˄R)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
(p∧q)⇔(p∧r)⇔(p∧(q⇔r))
$$\left(p \wedge q\right) ⇔ \left(p \wedge r\right) ⇔ \left(p \wedge \left(q ⇔ r\right)\right)$$
Solución detallada
$$q ⇔ r = \left(q \wedge r\right) \vee \left(\neg q \wedge \neg r\right)$$
$$p \wedge \left(q ⇔ r\right) = p \wedge \left(q \vee \neg r\right) \wedge \left(r \vee \neg q\right)$$
$$\left(p \wedge q\right) ⇔ \left(p \wedge r\right) ⇔ \left(p \wedge \left(q ⇔ r\right)\right) = \left(q \wedge r\right) \vee \neg p$$
$$p \wedge \left(q ⇔ r\right) = p \wedge \left(q \vee \neg r\right) \wedge \left(r \vee \neg q\right)$$
$$\left(p \wedge q\right) ⇔ \left(p \wedge r\right) ⇔ \left(p \wedge \left(q ⇔ r\right)\right) = \left(q \wedge r\right) \vee \neg p$$
Tabla de verdad
+---+---+---+--------+ | p | q | r | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 0 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+