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| 1 | a) Disjunkte Mengen bilden ein Gegenbeispiel: | ||
| 2 | |||
| 3 | \begin{align*} | ||
| 4 | A &= \{ a \} \\ | ||
| 5 | B &= \{ b \} \\ | ||
| 6 | C &= \{ c \} \\ | ||
| 7 | D &= \{ d \} \\ | ||
| 8 | A - (B - (C - D)) &= \{ a \} \\ | ||
| 9 | (A \cup C) - (B \cup D) &= \{ a, c \} | ||
| 10 | \end{align*} | ||
| 11 | |||
| 12 | <!-- Es sei $\Omega = A \cup B \cup C \cup D$ --> | ||
| 13 | |||
| 14 | <!-- \begin{align*} --> | ||
| 15 | <!-- \ & A - (B - (C - D)) = (A \cup C) - (B \cup D) \\ --> | ||
| 16 | <!-- \underset{\text{Extens.}}{\lequiv}& \forall x \in \Omega .\ \iota_A(x) \land \lnot (\iota_B(x) \land \lnot (\iota_C(x) \land \lnot \iota_D(x))) \lequiv (\iota_A(x) \lor \iota_C(x)) \land \lnot (\iota_B(x) \lor \iota_D(x)) \\ --> | ||
| 17 | <!-- \lequiv& \forall x \in \Omega .\ \iota_A(x) \land (\lnot \iota_B(x) \lor (\iota_C(x) \land \lnot \iota_D(x))) \lequiv (\iota_A(x) \lor \iota_C(x)) \land (\lnot \iota_B(x) \land \lnot \iota_D(x)) \\ --> | ||
| 18 | <!-- \lequiv& \forall x \in \Omega .\ (\iota_A(x) \land \lnot \iota_B(x)) \lor (\iota_A(x) \land \iota_C(x) \land \lnot \iota_D(x))) \lequiv (\iota_A(x) \land \lnot \iota_B(x) \land \lnot \iota_D(x)) \lor (\lnot \iota_B(x) \land \iota_C(x) \land \lnot \iota_D(x)) --> | ||
| 19 | <!-- \end{align*} --> | ||
| 20 | |||
| 21 | b) Es sei $\Omega = A \cup B$ und für eine Menge $X$ sei $\bar X = \Omega - X$. | ||
| 22 | $\Omega$ bildet mit $\cap, \cup$ und dem soeben definierten Komplement eine boolsche Algebra. | ||
| 23 | |||
| 24 | Die Differenz $A - B$ ist definiert als $A \cap \bar B$. | ||
| 25 | |||
| 26 | \begin{align*} | ||
| 27 | (A - B) \cup (B - A) &\underset{\text{\scriptsize Diff.}}{=} (A \cap \bar B) \cup (B \cap \bar A) \\ | ||
| 28 | &\underset{\text{\scriptsize Dist.}}{=} ((A \cap \bar B) \cup B) \cap ((A \cap \bar B) \cup \bar A) \\ | ||
| 29 | &\underset{\text{\scriptsize Dist.}}{=} ((A \cup B) \cap (\bar B \cup B)) \cap ((A \cup \bar A) \cap (\bar B \cup \bar A)) \\ | ||
| 30 | &\underset{\text{\scriptsize Tnd.\footnotemark}}{=} (A \cup B) \cap (\bar A \cup \bar B) \\ | ||
| 31 | &\underset{\text{\scriptsize Diff.}}{=} (A \cup B) - (A \cap B) | ||
| 32 | \end{align*} | ||
| 33 | \footnotetext{Tertium non datur} | ||
