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