Sr Examen
Expresión ¬(P⇔Q)
⌨
Solución
Solución detallada
$$p ⇔ q = \left(p \wedge q\right) \vee \left(\neg p \wedge \neg q\right)$$
$$p \not\equiv q = \left(p \wedge \neg q\right) \vee \left(q \wedge \neg p\right)$$
$$p \not\equiv q = \left(p \wedge \neg q\right) \vee \left(q \wedge \neg p\right)$$
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 | +---+---+--------+
FND
[src]
Ya está reducido a FND
$$\left(p \wedge \neg q\right) \vee \left(q \wedge \neg p\right)$$
(p∧(¬q))∨(q∧(¬p))
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))