Sr Examen
Expresión ¬q→(p∧r)≡(¬q→r)∧(q∨p)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
((¬q)⇒(p∧r))⇔((p∨q)∧((¬q)⇒r))
$$\left(\left(\neg q \Rightarrow r\right) \wedge \left(p \vee q\right)\right) ⇔ \left(\neg q \Rightarrow \left(p \wedge r\right)\right)$$
Solución detallada
$$\neg q \Rightarrow \left(p \wedge r\right) = q \vee \left(p \wedge r\right)$$
$$\neg q \Rightarrow r = q \vee r$$
$$\left(\neg q \Rightarrow r\right) \wedge \left(p \vee q\right) = q \vee \left(p \wedge r\right)$$
$$\left(\left(\neg q \Rightarrow r\right) \wedge \left(p \vee q\right)\right) ⇔ \left(\neg q \Rightarrow \left(p \wedge r\right)\right) = 1$$
$$\neg q \Rightarrow r = q \vee r$$
$$\left(\neg q \Rightarrow r\right) \wedge \left(p \vee q\right) = q \vee \left(p \wedge r\right)$$
$$\left(\left(\neg q \Rightarrow r\right) \wedge \left(p \vee q\right)\right) ⇔ \left(\neg q \Rightarrow \left(p \wedge r\right)\right) = 1$$
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 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+