Sr Examen
Expresión ((~P∧Q)→(R∧~R))∧~Q
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Solución
Ha introducido
[src]
(¬q)∧((q∧(¬p))⇒(r∧(¬r)))
$$\left(\left(q \wedge \neg p\right) \Rightarrow \left(r \wedge \neg r\right)\right) \wedge \neg q$$
Solución detallada
$$r \wedge \neg r = \text{False}$$
$$\left(q \wedge \neg p\right) \Rightarrow \left(r \wedge \neg r\right) = p \vee \neg q$$
$$\left(\left(q \wedge \neg p\right) \Rightarrow \left(r \wedge \neg r\right)\right) \wedge \neg q = \neg q$$
$$\left(q \wedge \neg p\right) \Rightarrow \left(r \wedge \neg r\right) = p \vee \neg q$$
$$\left(\left(q \wedge \neg p\right) \Rightarrow \left(r \wedge \neg r\right)\right) \wedge \neg q = \neg q$$
Tabla de verdad
+---+---+---+--------+ | p | q | r | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+