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Expresión ((P∨¬Q)∧(¬Q∨P)∧(¬Q∨R))→(P∨¬Q)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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