Sr Examen

Expresión b(((¬(ab)⇒c)⇒b)⇒(a¬c⇔a+b))

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    Solución

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