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Expresión A→(B∨C)≡(A→B)∨(A→C)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

    Ha introducido [src]
    (a⇒(b∨c))⇔((a⇒b)∨(a⇒c))
    $$\left(a \Rightarrow \left(b \vee c\right)\right) ⇔ \left(\left(a \Rightarrow b\right) \vee \left(a \Rightarrow c\right)\right)$$
    Solución detallada
    $$a \Rightarrow \left(b \vee c\right) = b \vee c \vee \neg a$$
    $$a \Rightarrow b = b \vee \neg a$$
    $$a \Rightarrow c = c \vee \neg a$$
    $$\left(a \Rightarrow b\right) \vee \left(a \Rightarrow c\right) = b \vee c \vee \neg a$$
    $$\left(a \Rightarrow \left(b \vee c\right)\right) ⇔ \left(\left(a \Rightarrow b\right) \vee \left(a \Rightarrow c\right)\right) = 1$$
    Simplificación [src]
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    1
    Tabla de verdad
    +---+---+---+--------+
    | a | b | c | 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
    FNCD [src]
    1
    1
    FNDP [src]
    1
    1