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