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