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