Sr Examen

Expresión (P∨Q∨R)∧(P∨T∨¬Q)∧(P∨¬T∨R)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    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))
    FNDP [src]
    $$p \vee \left(r \wedge t\right) \vee \left(r \wedge \neg q\right)$$
    p∨(r∧t)∨(r∧(¬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))
    FNCD [src]
    $$\left(p \vee r\right) \wedge \left(p \vee t \vee \neg q\right)$$
    (p∨r)∧(p∨t∨(¬q))