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