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