Sr Examen

Expresión pv(q&(pv(¬r)))

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

    Ha introducido [src]
    p∨(q∧(p∨(¬r)))
    $$p \vee \left(q \wedge \left(p \vee \neg r\right)\right)$$
    Solución detallada
    $$p \vee \left(q \wedge \left(p \vee \neg r\right)\right) = p \vee \left(q \wedge \neg r\right)$$
    Simplificación [src]
    $$p \vee \left(q \wedge \neg r\right)$$
    p∨(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 | 1      |
    +---+---+---+--------+
    FNCD [src]
    $$\left(p \vee q\right) \wedge \left(p \vee \neg r\right)$$
    (p∨q)∧(p∨(¬r))
    FND [src]
    Ya está reducido a FND
    $$p \vee \left(q \wedge \neg r\right)$$
    p∨(q∧(¬r))
    FNC [src]
    $$\left(p \vee q\right) \wedge \left(p \vee \neg r\right)$$
    (p∨q)∧(p∨(¬r))
    FNDP [src]
    $$p \vee \left(q \wedge \neg r\right)$$
    p∨(q∧(¬r))