Exercice :
Soient \(A\), \(B\) et \(C\) trois parties d'un ensemble \(E\).
Donner une écriture simplifiée des sous-ensembles suivants :
1) \([A \cup (A \cap B)] \cap B\).
2) \((A \cap B) \cup \left(A \cap \overline{B}\right)\).
3) \(\overline{(A \cup B)} \cap \left(C \cup \overline{A}\right)\).
4) \([(A \cup B) \cap (B \cap C)] \cup (A \cup C)\).
5) \((A \cup B) \cap [(B \cap C) \cup (A \cup C)]\).
6) \(B \cup (A \setminus B)\).
7) \((A \cap B) \cup (A \cap \overline{B})
\cup (\overline{A} \cap B)
\cup (\overline{A} \cap \overline{B})\).
8) \([(A \cap B) \cup (A \cup \overline{B})
\cup (\overline{A} \cap B)]
\cap (\overline{A} \cup \overline{B})\).
Corrigé :
1) Puisque \(A \cap B \subset A\), alors
\[
A \cup (A \cap B)=A.
\]
Donc
\[
[A \cup (A \cap B)] \cap B=A \cap B.
\]
2) On a :
\[
\begin{aligned}
(A \cap B) \cup (A \cap \overline{B})
&= A \cap (B \cup \overline{B}) \\
&= A \cap E \\
&= A.
\end{aligned}
\]
3) On a :
\[
\begin{aligned}
\overline{(A \cup B)} \cap (C \cup \overline{A})
&= (\overline{A} \cap \overline{B}) \cap (C \cup \overline{A}).
\end{aligned}
\]
Or,
\[
\overline{A} \cap \overline{B}
\subset \overline{A}
\subset C \cup \overline{A}.
\]
Donc
\[
(\overline{A} \cap \overline{B}) \cap
(C \cup \overline{A})
= \overline{A} \cap \overline{B}.
\]
D'où
\[
\overline{(A \cup B)} \cap (C \cup \overline{A})
= \overline{A \cup B}.
\]
4) On a :
\[
B \cap C \subset B \subset A \cup B.
\]
Donc
\[
(A \cup B) \cap (B \cap C)=B \cap C.
\]
Ainsi,
\[
\begin{aligned}
&[(A \cup B) \cap (B \cap C)] \cup (A \cup C)\\
&\qquad=(B \cap C) \cup (A \cup C).
\end{aligned}
\]
Or,
\[
B \cap C \subset C \subset A \cup C.
\]
Donc
\[
(B \cap C) \cup (A \cup C)=A \cup C.
\]
D'où
\[
[(A \cup B) \cap (B \cap C)] \cup (A \cup C)
= A \cup C.
\]
5) On a :
\[
B \cap C \subset C \subset A \cup C.
\]
Donc
\[
(B \cap C) \cup (A \cup C)=A \cup C.
\]
D'où
\[
\begin{aligned}
(A \cup B) \cap [(B \cap C) \cup (A \cup C)]
&= (A \cup B) \cap (A \cup C) \\
&= A \cup (B \cap C).
\end{aligned}
\]
6) On a :
\[
\begin{aligned}
B \cup (A \cap \overline{B})
&= (B \cup A) \cap (B \cup \overline{B}) \\
&= (B \cup A) \cap E \\
&= B \cup A.
\end{aligned}
\]
7) On a :
\[
\begin{aligned}
(A \cap B) \cup (A \cap \overline{B})
&= A \cap (B \cup \overline{B}) \\
&= A \cap E \\
&= A.
\end{aligned}
\]
De même, on a :
\[
\begin{aligned}
(\overline{A} \cap B) \cup
(\overline{A} \cap \overline{B})
&= \overline{A} \cap (B \cup \overline{B}) \\
&= \overline{A} \cap E \\
&= \overline{A}.
\end{aligned}
\]
Donc
\[
\begin{aligned}
&(A \cap B) \cup (A \cap \overline{B})
\cup (\overline{A} \cap B)
\cup (\overline{A} \cap \overline{B}) \\
&\qquad = A \cup \overline{A} \\
&\qquad = E.
\end{aligned}
\]
8) On a :
\[
\begin{aligned}
(A \cap B) \cup (\overline{A} \cap B)
&= (A \cup \overline{A}) \cap B \\
&= E \cap B \\
&= B.
\end{aligned}
\]
Donc
\[
\begin{aligned}
&(A \cap B) \cup (A \cup \overline{B})
\cup (\overline{A} \cap B) \\
&\quad = [(A \cap B) \cup (\overline{A} \cap B)]
\cup (A \cup \overline{B}) \\
&\quad = B \cup (A \cup \overline{B}) \\
&\quad = (B \cup \overline{B}) \cup A \\
&\quad = E \cup A \\
&\quad = E.
\end{aligned}
\]
D'où
\[
\begin{aligned}
&[(A \cap B) \cup (A \cup \overline{B})
\cup (\overline{A} \cap B)]
\cap (\overline{A} \cup \overline{B}) \\
&\quad = E \cap (\overline{A} \cup \overline{B}) \\
&\quad = \overline{A} \cup \overline{B} \\
&\quad = \overline{A \cap B}.
\end{aligned}
\]