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