Sr Examen

Expresión ¬((¬p)∨¬q)→(r∨¬t)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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