Sr Examen

Expresión x∨(y⇒z)≡(x∨y)⇒(x∨z)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

    Ha introducido [src]
    (x∨(y⇒z))⇔((x∨y)⇒(x∨z))
    $$\left(\left(x \vee y\right) \Rightarrow \left(x \vee z\right)\right) ⇔ \left(x \vee \left(y \Rightarrow z\right)\right)$$
    Solución detallada
    $$y \Rightarrow z = z \vee \neg y$$
    $$x \vee \left(y \Rightarrow z\right) = x \vee z \vee \neg y$$
    $$\left(x \vee y\right) \Rightarrow \left(x \vee z\right) = x \vee z \vee \neg y$$
    $$\left(\left(x \vee y\right) \Rightarrow \left(x \vee z\right)\right) ⇔ \left(x \vee \left(y \Rightarrow z\right)\right) = 1$$
    Simplificación [src]
    1
    1
    Tabla de verdad
    +---+---+---+--------+
    | x | y | z | 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      |
    +---+---+---+--------+
    FNCD [src]
    1
    1
    FND [src]
    Ya está reducido a FND
    1
    1
    FNC [src]
    Ya está reducido a FNC
    1
    1
    FNDP [src]
    1
    1