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