Sr Examen

Expresión ¬(P→(Q∧R))

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

    Ha introducido [src]
    ¬(p⇒(q∧r))
    $$p \not\Rightarrow \left(q \wedge r\right)$$
    Solución detallada
    $$p \Rightarrow \left(q \wedge r\right) = \left(q \wedge r\right) \vee \neg p$$
    $$p \not\Rightarrow \left(q \wedge r\right) = p \wedge \left(\neg q \vee \neg r\right)$$
    Simplificación [src]
    $$p \wedge \left(\neg q \vee \neg r\right)$$
    p∧((¬q)∨(¬r))
    Tabla de verdad
    +---+---+---+--------+
    | p | q | r | result |
    +===+===+===+========+
    | 0 | 0 | 0 | 0      |
    +---+---+---+--------+
    | 0 | 0 | 1 | 0      |
    +---+---+---+--------+
    | 0 | 1 | 0 | 0      |
    +---+---+---+--------+
    | 0 | 1 | 1 | 0      |
    +---+---+---+--------+
    | 1 | 0 | 0 | 1      |
    +---+---+---+--------+
    | 1 | 0 | 1 | 1      |
    +---+---+---+--------+
    | 1 | 1 | 0 | 1      |
    +---+---+---+--------+
    | 1 | 1 | 1 | 0      |
    +---+---+---+--------+
    FNDP [src]
    $$\left(p \wedge \neg q\right) \vee \left(p \wedge \neg r\right)$$
    (p∧(¬q))∨(p∧(¬r))
    FNC [src]
    Ya está reducido a FNC
    $$p \wedge \left(\neg q \vee \neg r\right)$$
    p∧((¬q)∨(¬r))
    FNCD [src]
    $$p \wedge \left(\neg q \vee \neg r\right)$$
    p∧((¬q)∨(¬r))
    FND [src]
    $$\left(p \wedge \neg q\right) \vee \left(p \wedge \neg r\right)$$
    (p∧(¬q))∨(p∧(¬r))