Erklärungen zur Differenzmenge (Youtube-Video der KhanAcademyDeutsch ) Mit Hilfe der Differenz kannst du alle Elemente einer Menge
A
{\displaystyle A}
B
{\displaystyle B}
x
{\displaystyle x}
A
{\displaystyle A}
B
{\displaystyle B}
x
{\displaystyle x}
A
{\displaystyle A}
B
{\displaystyle B}
Differenz zweier Mengen Für die Differenz von
A
{\displaystyle A}
B
{\displaystyle B}
A
∖
B
{\displaystyle A\setminus B}
A
{\displaystyle A}
B
{\displaystyle B}
A
−
B
{\displaystyle A-B}
Beispiel (Differenz)
{
2
,
4
,
5
,
9
}
∖
{
1
,
3
,
5
,
7
,
12
}
=
{
2
,
4
,
9
}
{\displaystyle \{2,4,5,9\}\setminus \{1,3,5,7,12\}=\{2,4,9\}}
{
3
,
12
,
18
}
∖
{
1
,
5
,
7
}
=
{
3
,
12
,
18
}
{\displaystyle \{3,12,18\}\setminus \{1,5,7\}=\{3,12,18\}}
{
x
∈
N
|
x
gerade
}
∖
{
x
∈
N
|
x
ungerade
}
=
{
x
∈
N
|
x
gerade
}
{\displaystyle \{x\in \mathbb {N} |\,x{\text{ gerade}}\}\setminus \{x\in \mathbb {N} |\,x{\text{ ungerade}}\}=\{x\in \mathbb {N} |\,x{\text{ gerade}}\}}
symmetrische Differenz zweier Mengen Beispiel für die symmetrische Differenz von zwei Mengen Die Symmetrische Differenz zweier Mengen
A
{\displaystyle A}
B
{\displaystyle B}
genau einer der Mengen
A
{\displaystyle A}
B
{\displaystyle B}
x
{\displaystyle x}
A
{\displaystyle A}
B
{\displaystyle B}
x
∈
A
{\displaystyle x\in A}
x
∈
B
{\displaystyle x\in B}
x
{\displaystyle x}
A
{\displaystyle A}
B
{\displaystyle B}
A
△
B
{\displaystyle A\,\triangle \,B}
A
{\displaystyle A}
B
{\displaystyle B}
Verständnisfrage: Bilde die symmetrische Differenz folgender Mengen
{
1
,
2
,
3
}
△
{
2
,
3
,
4
}
=
?
{\displaystyle \{1,2,3\}\,\triangle \,\{2,3,4\}=?}
{
1
,
2
,
3
}
△
{
1
,
2
,
3
}
=
?
{\displaystyle \{1,2,3\}\,\triangle \,\{1,2,3\}=?}
{
1
,
2
,
3
}
△
{
4
,
5
,
6
}
=
?
{\displaystyle \{1,2,3\}\,\triangle \,\{4,5,6\}=?}
Z
△
R
=
?
{\displaystyle \mathbb {Z} \,\triangle \,\mathbb {R} =?}
N
△
∅
=
?
{\displaystyle \mathbb {N} \,\triangle \,\varnothing =?}
Antwort:
{
1
,
2
,
3
}
△
{
2
,
3
,
4
}
=
{
1
,
4
}
{\displaystyle \{1,2,3\}\,\triangle \,\{2,3,4\}=\{1,4\}}
{
1
,
2
,
3
}
△
{
1
,
2
,
3
}
=
∅
{\displaystyle \{1,2,3\}\,\triangle \,\{1,2,3\}=\emptyset }
{
1
,
2
,
3
}
△
{
4
,
5
,
6
}
=
{
1
,
2
,
3
,
4
,
5
,
6
}
{\displaystyle \{1,2,3\}\,\triangle \,\{4,5,6\}=\{1,2,3,4,5,6\}}
Z
△
R
=
R
∖
Z
=
{
x
∈
R
|
x
ist keine ganze Zahl.
}
{\displaystyle \mathbb {Z} \,\triangle \,\mathbb {R} =\mathbb {R} \setminus \mathbb {Z} =\left\{x\in \mathbb {R} \,|\,x{\text{ ist keine ganze Zahl.}}\right\}}
N
△
∅
=
N
{\displaystyle \mathbb {N} \,\triangle \,\varnothing =\mathbb {N} }
das Komplement einer Menge Erklärungen zur Grundmenge und dem absoluten Komplement (Youtube-Video der KhanAcademyDeutsch ) Das Komplement einer Menge
A
{\displaystyle A}
A
{\displaystyle A}
R
{\displaystyle \mathbb {R} }
Allklasse
V
{\displaystyle {\mathcal {V}}}
A
{\displaystyle A}
A
∁
{\displaystyle A^{\complement }}
∁
A
{\displaystyle \complement A}
A
¯
{\displaystyle {\overline {A}}}
A
{\displaystyle A}
Antwort:
{
0
,
1
,
2
,
3
}
∁
=
N
≥
4
{\displaystyle \{0,1,2,3\}^{\complement }=\mathbb {N} _{\geq 4}}
{
x
∈
N
0
|
x
ist gerade
}
∁
=
{
x
∈
N
0
|
x
ist ungerade
}
{\displaystyle \left\{x\in \mathbb {N} _{0}\,|\,x{\text{ ist gerade}}\right\}^{\complement }=\left\{x\in \mathbb {N} _{0}\,|\,x{\text{ ist ungerade}}\right\}}
N
0
∁
=
∅
{\displaystyle \mathbb {N} _{0}^{\complement }=\varnothing }
∅
∁
=
N
0
{\displaystyle \varnothing ^{\complement }=\mathbb {N} _{0}}
Dieses Eigenschaften lassen sich auf die Definitionen und die entsprechenden Gesetze der Logik zurückführen.
Beweis (Eigenschaften des Komplements)
Es ist
(
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∁
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∁
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}
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∉
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∉
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}
=
{
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|
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∈
A
}
↕
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a
¨
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p
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=
A
{\displaystyle {\begin{aligned}&(A^{\complement })^{\complement }\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{y\,|\,y\notin A\}^{\complement }\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\,|\,x\notin \{y\,|\,y\notin A\}\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin \{y\,|\,y\notin A\}}}\right.}\\[0.3em]=\;&\{x\,|\,x\in A\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Extensionalit{\ddot {a}}tsprinzip}}\right.}\\[0.3em]=\;&A\end{aligned}}}
Es ist
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∈
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∧
x
∈
{
y
|
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∉
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}
}
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}
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d
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K
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≠
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}
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∅
=
∅
{\displaystyle {\begin{aligned}&A\cap A^{\complement }\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&A\cap \{y|\,y\notin A\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\cap }}\right.}\\[0.3em]=\;&\{x\;|\;x\in A\land x\in \{y|\,y\notin A\}\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin \{y\,|\,y\notin A\}}}\right.}\\[0.3em]=\;&\{x\;|\;x\in A\land x\notin A\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {die\;Konjunktion\;ist\;immer\;falsch}}\right.}\\[0.3em]=\;&\{x\;|\;x\neq x\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\varnothing }}\right.}\\[0.3em]=\;&\varnothing \end{aligned}}}
Siehe Aufgabe 1
Es ist
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∈
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∧
y
∈
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}
=
{
x
|
¬
(
x
∈
A
∧
x
∈
B
)
}
↕
d
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∨
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}
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∈
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∉
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}
∨
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∈
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}
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∁
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∩
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A
∁
∪
B
∁
{\displaystyle {\begin{aligned}&(A\cap B)^{\complement }\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\;|\;x\notin A\cap B\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\cap }}\right.}\\[0.3em]=\;&\{x\;|\;x\notin \{y\;|\;y\in A\land y\in B\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin \{y\,|\,y\in A\land y\in B\}}}\right.}\\[0.3em]=\;&\{x\;|\;\neg \,(x\in A\land x\in B)\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {de\;Morgansche\;Regel\;f{\ddot {u}}r\;\neg \;und\;\land }}\right.}\\[0.3em]=\;&\{x\;|\;x\notin A\lor x\notin B\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin A\;und\;x\notin B}}\right.}\\[0.3em]=\;&\{x\;|\;x\in \{y\,|y\notin A\}\lor x\in \{z\,|\,z\notin B\}\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\;|\;x\in A^{\complement }\lor x\in B^{\complement }\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\cap }}\right.}\\[0.3em]=\;&A^{\complement }\cup B^{\complement }\end{aligned}}}
Siehe Aufgabe 2
Es ist
V
∁
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{\displaystyle {\begin{aligned}&{\mathcal {V}}^{\complement }\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\,|\,x\notin {\mathcal {V}}\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ x\notin {\mathcal {V}}\;{\mathsf {ist\;immer\;falsch}}\right.}\\[0.3em]=\;&\{x\,|\,x\neq x\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\varnothing }}\right.}\\[0.3em]=\;&\varnothing \end{aligned}}}
Siehe Aufgabe 3
Aufgabe (Aufgabe 1)
Zeige:
A
∪
A
∁
=
V
{\displaystyle A\cup A^{\complement }={\mathcal {V}}}
Lösung (Aufgabe 1)
A
∪
A
∁
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}
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∈
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∨
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∈
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∉
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}
}
↕
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=
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}
↕
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{\displaystyle {\begin{aligned}&A\cup A^{\complement }\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&A\cup \{y|\,y\notin A\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\cup }}\right.}\\[0.3em]=\;&\{x\;|\;x\in A\lor x\in \{y|\,y\notin A\}\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin \{y\,|\,y\notin A\}}}\right.}\\[0.3em]=\;&\{x\;|\;x\in A\lor x\notin A\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {die\;Disjunktion\;ist\;immer\;wahr}}\right.}\\[0.3em]=\;&\{x\;|\;x=x\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;{\mathcal {V}}}}\right.}\\[0.3em]=\;&{\mathcal {V}}\end{aligned}}}
Aufgabe (Aufgabe 2)
Zeige:
(
A
∪
B
)
∁
=
A
∁
∩
B
∁
{\displaystyle (A\cup B)^{\complement }=A^{\complement }\cap B^{\complement }}
Lösung (Aufgabe 2)
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∪
B
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∁
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∈
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}
↕
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=
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}
↕
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{\displaystyle {\begin{aligned}&(A\cup B)^{\complement }\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\;|\;x\notin A\cup B\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\cup }}\right.}\\[0.3em]=\;&\{x\;|\;x\notin \{y\;|\;y\in A\lor y\in B\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin \{y\,|\,y\in A\lor y\in B\}}}\right.}\\[0.3em]=\;&\{x\;|\;\neg \,(x\in A\lor x\in B)\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {de\;Morgansche\;Regel\;f{\ddot {u}}r\;\neg \;und\;\lor }}\right.}\\[0.3em]=\;&\{x\;|\;x\notin A\land x\notin B\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin A\;und\;x\notin B}}\right.}\\[0.3em]=\;&\{x\;|\;x\in \{y\,|y\notin A\}\land x\in \{z\,|\,z\notin B\}\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\;|\;x\in A^{\complement }\land x\in B^{\complement }\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\cap }}\right.}\\[0.3em]=\;&A^{\complement }\cap B^{\complement }\end{aligned}}}
Aufgabe 3: Zeige:
∅
∁
=
V
{\displaystyle \varnothing ^{\complement }={\mathcal {V}}}
∅
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V
{\displaystyle {\begin{aligned}&\varnothing ^{\complement }\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\,|\,x\notin \varnothing \}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ x\notin \varnothing \;{\mathsf {ist\;immer\;wahr}}\right.}\\[0.3em]=\;&\{x\,|\,x=x\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;{\mathcal {V}}}}\right.}\\[0.3em]=\;&{\mathcal {V}}\end{aligned}}}
![{\displaystyle {\begin{aligned}A\,\triangle \,B&:=(A\setminus B)\cup (B\setminus A)\\[0.3em]&=\{x\in A\cup B|\,(x\in A\land x\notin B)\lor (x\notin A\land x\in B)\}\\[0.3em]&=\{x\in A\cup B|x\in A\,{\dot {\lor }}\,x\in B\}\\[0.3em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12d66b766ce3a0e8c79a9b2e42d4a9a2744e973e)
![{\displaystyle {\begin{aligned}&(A^{\complement })^{\complement }\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{y\,|\,y\notin A\}^{\complement }\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\,|\,x\notin \{y\,|\,y\notin A\}\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin \{y\,|\,y\notin A\}}}\right.}\\[0.3em]=\;&\{x\,|\,x\in A\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Extensionalit{\ddot {a}}tsprinzip}}\right.}\\[0.3em]=\;&A\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc6efa930685485ab8db74d8be5f0b56ff6a9e18)
![{\displaystyle {\begin{aligned}&A\cap A^{\complement }\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&A\cap \{y|\,y\notin A\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\cap }}\right.}\\[0.3em]=\;&\{x\;|\;x\in A\land x\in \{y|\,y\notin A\}\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin \{y\,|\,y\notin A\}}}\right.}\\[0.3em]=\;&\{x\;|\;x\in A\land x\notin A\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {die\;Konjunktion\;ist\;immer\;falsch}}\right.}\\[0.3em]=\;&\{x\;|\;x\neq x\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\varnothing }}\right.}\\[0.3em]=\;&\varnothing \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0be2f6da1653e4614e9c9d382240948b0b376262)
![{\displaystyle {\begin{aligned}&(A\cap B)^{\complement }\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\;|\;x\notin A\cap B\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\cap }}\right.}\\[0.3em]=\;&\{x\;|\;x\notin \{y\;|\;y\in A\land y\in B\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin \{y\,|\,y\in A\land y\in B\}}}\right.}\\[0.3em]=\;&\{x\;|\;\neg \,(x\in A\land x\in B)\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {de\;Morgansche\;Regel\;f{\ddot {u}}r\;\neg \;und\;\land }}\right.}\\[0.3em]=\;&\{x\;|\;x\notin A\lor x\notin B\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin A\;und\;x\notin B}}\right.}\\[0.3em]=\;&\{x\;|\;x\in \{y\,|y\notin A\}\lor x\in \{z\,|\,z\notin B\}\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\;|\;x\in A^{\complement }\lor x\in B^{\complement }\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\cap }}\right.}\\[0.3em]=\;&A^{\complement }\cup B^{\complement }\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23ab913adb087c36973f40c286f14b06bc54cd74)
![{\displaystyle {\begin{aligned}&{\mathcal {V}}^{\complement }\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\,|\,x\notin {\mathcal {V}}\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ x\notin {\mathcal {V}}\;{\mathsf {ist\;immer\;falsch}}\right.}\\[0.3em]=\;&\{x\,|\,x\neq x\}\\[0.3em]&{\color {OliveGreen}\left\updownarrow \ {\mathsf {Definition\;von\;\varnothing }}\right.}\\[0.3em]=\;&\varnothing \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69108fa6ec83830c5db17eec4bf76620531ef635)
![{\displaystyle {\begin{aligned}&A\cup A^{\complement }\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&A\cup \{y|\,y\notin A\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\cup }}\right.}\\[0.3em]=\;&\{x\;|\;x\in A\lor x\in \{y|\,y\notin A\}\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin \{y\,|\,y\notin A\}}}\right.}\\[0.3em]=\;&\{x\;|\;x\in A\lor x\notin A\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {die\;Disjunktion\;ist\;immer\;wahr}}\right.}\\[0.3em]=\;&\{x\;|\;x=x\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;{\mathcal {V}}}}\right.}\\[0.3em]=\;&{\mathcal {V}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5cd37270437cfae5d8c79aa9d7087e31226b60f)
![{\displaystyle {\begin{aligned}&(A\cup B)^{\complement }\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\;|\;x\notin A\cup B\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\cup }}\right.}\\[0.3em]=\;&\{x\;|\;x\notin \{y\;|\;y\in A\lor y\in B\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin \{y\,|\,y\in A\lor y\in B\}}}\right.}\\[0.3em]=\;&\{x\;|\;\neg \,(x\in A\lor x\in B)\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {de\;Morgansche\;Regel\;f{\ddot {u}}r\;\neg \;und\;\lor }}\right.}\\[0.3em]=\;&\{x\;|\;x\notin A\land x\notin B\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Abstraktionsprinzip\;f{\ddot {u}}r\;x\notin A\;und\;x\notin B}}\right.}\\[0.3em]=\;&\{x\;|\;x\in \{y\,|y\notin A\}\land x\in \{z\,|\,z\notin B\}\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\;|\;x\in A^{\complement }\land x\in B^{\complement }\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\cap }}\right.}\\[0.3em]=\;&A^{\complement }\cap B^{\complement }\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/272df9ee1cc1da5983cd2da316b27752269b04a4)
![{\displaystyle {\begin{aligned}&\varnothing ^{\complement }\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;\complement }}\right.}\\[0.3em]=\;&\{x\,|\,x\notin \varnothing \}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ x\notin \varnothing \;{\mathsf {ist\;immer\;wahr}}\right.}\\[0.3em]=\;&\{x\,|\,x=x\}\\[0.3em]&{\color {NavyBlue}\left\updownarrow \ {\mathsf {Definition\;von\;{\mathcal {V}}}}\right.}\\[0.3em]=\;&{\mathcal {V}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d07080dee87429f50306eb522de107348aa38eed)
