Sr Examen
Expresión ¬(¬Q→¬P)∧(P→¬(P→R))
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
(p⇒(¬(p⇒r)))∧(¬((¬q)⇒(¬p)))
$$\left(p \Rightarrow p \not\Rightarrow r\right) \wedge \neg q \not\Rightarrow \neg p$$
Solución detallada
$$p \Rightarrow r = r \vee \neg p$$
$$p \not\Rightarrow r = p \wedge \neg r$$
$$p \Rightarrow p \not\Rightarrow r = \neg p \vee \neg r$$
$$\neg q \Rightarrow \neg p = q \vee \neg p$$
$$\neg q \not\Rightarrow \neg p = p \wedge \neg q$$
$$\left(p \Rightarrow p \not\Rightarrow r\right) \wedge \neg q \not\Rightarrow \neg p = p \wedge \neg q \wedge \neg r$$
$$p \not\Rightarrow r = p \wedge \neg r$$
$$p \Rightarrow p \not\Rightarrow r = \neg p \vee \neg r$$
$$\neg q \Rightarrow \neg p = q \vee \neg p$$
$$\neg q \not\Rightarrow \neg p = p \wedge \neg q$$
$$\left(p \Rightarrow p \not\Rightarrow r\right) \wedge \neg q \not\Rightarrow \neg p = p \wedge \neg q \wedge \neg r$$
Tabla de verdad
+---+---+---+--------+ | p | q | r | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 0 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+