Short Answer

Let S be a subset of a universal set U. The characteristic function of S is the function from U to the set {0,1} such that fs(x) = 1 if x belongs to S and if x does not belong to S and fs(x) = 0. Let A and B be sets. Show that for all $x \in U$
$\begin{aligned} (a) f_{A \cap B} (x) = f_{A} (x) \cdot f_{B} (x) \\ (b) f_{A \cup B} (x) = f_{A} (x) + f_{B} (x) - f_{A} (x) \cdot f_{B} (x) \\ (c) f_{\bar{A}} (x) = 1 - f_{A} (x) \\ (d) f_{A \oplus B} (x) = f_{A} (x) + f_{B} (x) - 2 f_{A} (x) \cdot f_{B} (x) \end{aligned}$

$\begin{aligned} (a) f_{A \cap B} (x) = f_{A} (x) \cdot f_{B} (x) \\ (b) f_{A \cup B} (x) = f_{A} (x) + f_{B} (x) - f_{A} (x) \cdot f_{B} (x) \\ (c) f_{\bar{A}} (x) = 1 - f_{A} (x) \\ (d) f_{A \oplus B} (x) = f_{A} (x) + f_{B} (x) - 2 f_{A} (x) \cdot f_{B} (x) \end{aligned}$

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Step: 1

$\begin{aligned} f_{A \cap B} (x) = 1 \\ \leftrightarrow x \in A \cap B \\ \leftrightarrow x \in Aand x \in B \\ \leftrightarrow f_{A} (x) = 1 and f_{B} (x) = 1 \\ \leftrightarrow f_{A} (x) \cdot f_{B} (x) = 1 \end{aligned}$

Step: 2

$\begin{aligned} f_{A \cup B} (x) = 1 \\ \leftrightarrow x \in Aor x \in B \\ \leftrightarrow f_{A} (x) = {lorf}_{B} (x) = 1 \\ \leftrightarrow f_{A} (x) + f_{B} (x) - f_{A} (x) \cdot f_{B} (x) = 1 \end{aligned}$

Step: 3

$\begin{aligned} f_{\bar{A}} (x) = 1 \\ \leftrightarrow x \in \bar{A} \\ \leftrightarrow x \notin A \\ \leftrightarrow f_{A} (x) = 0 \\ \leftrightarrow 1 - f_{A} (x) = 1 \end{aligned}$

$\begin{aligned} f_{A \oplus B} (x) = 1 \\ \leftrightarrow x \in A \oplus B \\ \leftrightarrow (x \in Aand x \notin B) or (x \notin Aand x \in B) \\ \leftrightarrow f_{A} (x) + f_{B} (x) - 2 f_{A} (x) \cdot f_{B} (x) = 1 \end{aligned}$

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