Sr Examen

Expresión Q→(P̅→(Q̅⋀A̅))

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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