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