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