Sr Examen

Expresión Pv¬QvR→PvQ&PvR

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

    Ha introducido [src]
    (p∨r∨(¬q))⇒(p∨r∨(p∧q))
    $$\left(p \vee r \vee \neg q\right) \Rightarrow \left(p \vee r \vee \left(p \wedge q\right)\right)$$
    Solución detallada
    $$p \vee r \vee \left(p \wedge q\right) = p \vee r$$
    $$\left(p \vee r \vee \neg q\right) \Rightarrow \left(p \vee r \vee \left(p \wedge q\right)\right) = p \vee q \vee r$$
    Simplificación [src]
    $$p \vee q \vee r$$
    p∨q∨r
    Tabla de verdad
    +---+---+---+--------+
    | p | q | r | result |
    +===+===+===+========+
    | 0 | 0 | 0 | 0      |
    +---+---+---+--------+
    | 0 | 0 | 1 | 1      |
    +---+---+---+--------+
    | 0 | 1 | 0 | 1      |
    +---+---+---+--------+
    | 0 | 1 | 1 | 1      |
    +---+---+---+--------+
    | 1 | 0 | 0 | 1      |
    +---+---+---+--------+
    | 1 | 0 | 1 | 1      |
    +---+---+---+--------+
    | 1 | 1 | 0 | 1      |
    +---+---+---+--------+
    | 1 | 1 | 1 | 1      |
    +---+---+---+--------+
    FNDP [src]
    $$p \vee q \vee r$$
    p∨q∨r
    FNCD [src]
    $$p \vee q \vee r$$
    p∨q∨r
    FND [src]
    Ya está reducido a FND
    $$p \vee q \vee r$$
    p∨q∨r
    FNC [src]
    Ya está reducido a FNC
    $$p \vee q \vee r$$
    p∨q∨r