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