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1a) Disjunkte Mengen bilden ein Gegenbeispiel:
2
3\begin{align*}
4A &= \{ a \} \\
5B &= \{ b \} \\
6C &= \{ c \} \\
7D &= \{ d \} \\
8A - (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
21b) 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
24Die 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}