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