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