Sr Examen
Expresión ¬(P⇒Q)∨¬(Q⇒R)
El profesor se sorprenderá mucho al ver tu solución correcta😉
⌨
Solución
Solución detallada
$$p \Rightarrow q = q \vee \neg p$$
$$p \not\Rightarrow q = p \wedge \neg q$$
$$q \Rightarrow r = r \vee \neg q$$
$$q \not\Rightarrow r = q \wedge \neg r$$
$$p \not\Rightarrow q \vee q \not\Rightarrow r = \left(p \wedge \neg q\right) \vee \left(q \wedge \neg r\right)$$
$$p \not\Rightarrow q = p \wedge \neg q$$
$$q \Rightarrow r = r \vee \neg q$$
$$q \not\Rightarrow r = q \wedge \neg r$$
$$p \not\Rightarrow q \vee q \not\Rightarrow r = \left(p \wedge \neg q\right) \vee \left(q \wedge \neg r\right)$$
Simplificación
[src]
$$\left(p \wedge \neg q\right) \vee \left(q \wedge \neg r\right)$$
(p∧(¬q))∨(q∧(¬r))
Tabla de verdad
+---+---+---+--------+ | p | q | r | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+
FNC
[src]
$$\left(p \vee q\right) \wedge \left(p \vee \neg r\right) \wedge \left(q \vee \neg q\right) \wedge \left(\neg q \vee \neg r\right)$$
(p∨q)∧(p∨(¬r))∧(q∨(¬q))∧((¬q)∨(¬r))
FND
[src]
Ya está reducido a FND
$$\left(p \wedge \neg q\right) \vee \left(q \wedge \neg r\right)$$
(p∧(¬q))∨(q∧(¬r))