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Expresión ab(¬a⇔a⇒¬a(b⇒a(¬(ab)⇔a)))

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    Solución

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