283 lines
11 KiB
TeX
283 lines
11 KiB
TeX
\documentclass[12pt]{article}
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\usepackage{ntnu}
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\usepackage{ntnu-math}
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\author{Øystein Tveit}
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\title{MA0301 Exercise 3}
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\begin{document}
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\ntnuTitle{}
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\break{}
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\begin{excs}
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\exc{}
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\begin{align*}
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\neg ((\neg p \wedge q) \vee (\neg p \wedge \neg q)) &\vee (p \wedge q) && \\
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\neg (\neg p \wedge (q \vee \neg q)) &\vee (p \wedge q) && \text{Distributive law} \\
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\neg (\neg p \wedge T) &\vee (p \wedge q) && \text{Complement law} \\
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\neg (\neg p) &\vee (p \wedge q) && \text{Identity law} \\
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p &\vee (p \wedge q) && \text{Double negation law} \\
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p & && \text{Absortion law} \\
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\end{align*}
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\exc{}
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\begin{align*}
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((p \wedge q) \vee (p \wedge \neg r) \vee \neg(\neg p \vee q)) &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \\
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((p \wedge q) \vee (p \wedge \neg r) \vee (\neg\neg p \wedge \neg q)) &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \text{De Morgans's law} \\
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((p \wedge q) \vee (p \wedge \neg r) \vee ( p \wedge \neg q)) &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \text{Double negation law} \\
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((p \wedge (q \vee \neg q)) \vee (p \wedge \neg r)) &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \text{Distributive law} \\
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((p \wedge T) \vee (p \wedge \neg r)) &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \text{Complement law} \\
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((p) \vee (p \wedge \neg r)) &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \text{Identity law} \\
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p &\vee ((r \vee s \vee \neg r) \wedge \neg q) && \text{Absortion law} \\
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p &\vee ((\neg q \wedge r) \vee (\neg q \wedge s) \vee (\neg q \wedge \neg r)) && \text{Distributive law} \\
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p &\vee (\neg q \wedge r) \vee (\neg q \wedge s) \vee (\neg q \wedge \neg r) && \text{Associative law}
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\end{align*}
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\exc{}
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\renewcommand{\theenumii}{\roman{enumii})}
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\renewcommand{\theenumiii}{\alph{enumiii})}
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\begin{subexcs}
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\subexc{}
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\begin{ssubexcs}
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\ssubexc{}
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\begin{gather*}
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\{\{2,3,5\} \cup \{6,4\}\} \cap \{4,6,8\} \\
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\{\{2,4,6\}\} \cap \{4,6,8\} \\
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\emptyset
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\end{gather*}
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\ssubexc{}
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\begin{align*}
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P(\{7,8,9\}) &- P(\{7,9\}) \\
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\{\{7,8,9\}, \{7,8\}, \{8,9\}, \{7,9\}, \{7\}, \{8\}, \{9\}, \emptyset\} &- \{\{7,9\}, \{7\}, \{9\}, \emptyset\} \\
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\{\{7,8,9\}, \{7,8\}, \{8,9\}, \{8\}\} & \\
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\end{align*}
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\ssubexc{}
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\begin{gather*}
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P(\emptyset) \\
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\{\emptyset\}
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\end{gather*}
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\ssubexc{}
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\begin{gather*}
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\{1, 3, 5\} \times \{0\} \\
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\{ \langle 1,0 \rangle, \langle 3,0 \rangle, \langle 5, 0 \rangle \}
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\end{gather*}
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\ssubexc{}
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\begin{gather*}
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\{2,4,6\} \times \emptyset \\
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\emptyset
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\end{gather*}
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\ssubexc{}
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\begin{gather*}
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P(\{0\}) \times P(\{1\}) \\
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\{\emptyset, \{0\}\} \times \{\emptyset, \{1\}\} \\
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\{\langle\emptyset,\emptyset\rangle, \langle\emptyset,\{1\}\rangle, \langle\{0\},\emptyset\rangle, \langle\{0\},\{1\}\rangle\}
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\end{gather*}
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\ssubexc{}
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\begin{gather*}
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P(P(\{2\})) \\
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P(\{\emptyset,\{2\}\}) \\
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\{ \{\{\emptyset\}, \{2\}\}, \{\{\emptyset\}\}, \{\{2\}\}, \emptyset \}
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\end{gather*}
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\end{ssubexcs}
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\subexc{}
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Because the elements in a power set can be represented as a binary tree where every leaf node is a set that has the cardinality of $1$, and that $\{\{x\} : x \in A\}$ would make up all the leaf nodes, we can reason that
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\[ |P(A) - \{\{x\} : x \in A\}| = \frac{n}{2} \]
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\end{subexcs}
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\renewcommand{\theenumii}{\alph{enumii})}
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\renewcommand{\theenumiii}{\roman{enumiii})}
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\exc{}
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\begin{subexcs}
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\subexc{}
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$\emptyset = \{\emptyset\}$ is {\color{red}False} because $|\emptyset| \neq |\{\emptyset\}|$
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\subexc{}
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$\emptyset = \{0\}$ is {\color{red}False} because $|\emptyset| \neq |\{0\}|$
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\subexc{}
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$|\emptyset| = 0$ is {\color{ForestGreen}True} because $\emptyset$ has $0$ elements
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\subexc{}
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$P(\emptyset)$ is {\color{red}False} because $P(\emptyset) = \{\{\emptyset\}\}$ has $1$ element
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\subexc{}
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$\emptyset = \{\}$ is {\color{ForestGreen}True} because the empty set is a subset of every possible set
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\subexc{}
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$\emptyset = \{x \in \mathbb{N} : x \leq 0 and x > 0\}$ is {\color{red}False} because $x \leq 0 \wedge x > 0 \equiv \F$, which means there are no such elements, and thus the set is empty
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\end{subexcs}
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\exc{}
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\begin{subexcs}
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\subexc{}
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\begin{align*}
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&A \cap (A \cup B) \\
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&\{x : x \in A \wedge x \in (A \cup B)\} \\
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&\{x : x \in A \wedge (x \in A \vee x \in B)\} \\
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&\{x : x \in A\} \\
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&A
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\end{align*}
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\subexc{}
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\begin{align*}
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&A-(B \cap C) \\
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&\{x : x \in A \wedge x \notin (B \cap C)\} \\
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&\{x : x \in A \wedge (x \notin B \wedge x \notin C)\} \\
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&\{x : (x \in A \wedge x \notin B) \vee (x \in A \wedge x \notin C)\} \\
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&\{x : x \in (A - B) \vee x \in (A - C)\} \\
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&\{x : x \in (A - B) \cup (A - C)\} \\
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&(A-B) \cup (A-C)
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\end{align*}
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\end{subexcs}
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\exc{}
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\renewcommand{\theenumii}{\roman{enumii})}
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\begin{subexcs}
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\subexc{}
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\begin{align*}
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&(A \cup B) \setminus (A \cap B) \\
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&\{x: x \in (A \cup B) \setminus (A \cap B)\} \\
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&\{x: x \in (A \cup B) \wedge x \notin (A \cap B)\} \\
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&\{x: (x \in A \vee x \in B) \wedge (x \notin A \vee x \notin B)\} \\
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&\{x: x \in A \wedge (x \notin A \vee x \notin B) \vee x \in B \wedge (x \notin A \vee x \notin B) \} \\
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&\{x: ((x \in A \wedge x \notin A) \vee (x \in A \wedge x \notin B)) \vee ((x \in B \wedge x \notin A) \vee (x \in B \wedge x \notin B)) \} \\
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&\{x: (F \vee (x \in A \wedge x \notin B)) \vee ((x \in B \wedge x \notin A) \vee F) \} \\
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&\{x: (x \in A \wedge x \notin B) \vee (x \in B \wedge x \notin A) \} \\
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&\{x: x \in (A - B) \vee x \in (B - A) \} \\
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&\{x: x \in (A - B) \cup (B - A) \} \\
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&(A - B) \cup (B - A) \\
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\end{align*}
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\subexc{}
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For this exercise, I counted the elements which was in either set but not both
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\[ A \Delta B = \{2, 4, 6, 7, 8\} \]
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\end{subexcs}
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\renewcommand{\theenumii}{\alph{enumii})}
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\exc{}
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\begin{align*}
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X &= \{\{1,2,3\}, \{2,3\}, \{ef\}\} \cup \{\{e\}\} \\
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&= \{\{1,2,3\}, \{2,3\}, \{ef\}, \{e\}\} \\
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\end{align*}
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\begin{align*}
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P(x) = \{ \\
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&\{\{1,2,3\}, \{2,3\}, \{ef\}, \{e\}\}, \\
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&\{\{1,2,3\}, \{2,3\}, \{ef\}\}, \\
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&\{\{1,2,3\}, \{2,3\}, \{e\}\}, \\
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&\{\{1,2,3\}, \{ef\}, \{e\}\}, \\
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&\{\{2,3\}, \{ef\}, \{e\}\}, \\
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&\{\{1,2,3\}, \{2,3\}\}, \\
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&\{\{1,2,3\}, \{e\}\}, \\
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&\{\{1,2,3\}, \{ef\}\}, \\
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&\{\{2,3\}, \{ef\}\}, \\
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&\{\{2,3\}, \{e\}\}, \\
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&\{\{ef\}, \{e\}\}, \\
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&\{\{e\}\}, \\
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&\{\{ef\}\}, \\
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&\{\{2,3\}\}, \\
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&\{\{1,2,3\}\} \\
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\} \\
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\end{align*}
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\begin{align*}
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P(X \cap Y) &= P(\{\{1,2,3\}, \{2,3\}, \{ef\}, \{e\}\} \cap \{\{1,2,3,e,f\}\}) \\
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&= P(\emptyset) \\
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&= \{\emptyset\}
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\end{align*}
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\exc{}
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\begin{subexcs}
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\subexc{}
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Here, the exercise says ``[\ldots] four sets $A_1$, $A_2$, $A_3$''. I'm not sure if I'm supposed to do three or four, but I'll assume three sets $A_1$, $A_2$, $A_3$ was the intention.
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\begin{align*}
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A_1 \cap A_2 \cap A_3 \\
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A_1 \cap A_2 \cap \overline{A_3} \\
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A_1 \cap \overline{A_2} \cap A_3 \\
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A_1 \cap \overline{A_2} \cap \overline{A_3} \\
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\overline{A_1} \cap A_2 \cap A_3 \\
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\overline{A_1} \cap A_2 \cap \overline{A_3} \\
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\overline{A_1} \cap \overline{A_2} \cap A_3 \\
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\overline{A_1} \cap \overline{A_2} \cap \overline{A_3} \\
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\end{align*}
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\subexc{}
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For each set, the amount of fundamental products is multiplied by $2$. Therefore, the amount of fundamental sets of $n$ sets is $2^n$
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\end{subexcs}
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\exc{}
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\begin{gather*}
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A \overline{( B \overline{C} )} \overline{( (A \overline{B}) \overline{C})} \\
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A ( \overline{B} + \overline{\overline{C}} ) \overline{(A \overline{B}\ \overline{C})} \\
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A ( \overline{B} + C ) \overline{(A \overline{B}\ \overline{C})} \\
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A ( \overline{B} + C ) \overline{(A \overline{B}\ \overline{C})} \\
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A ( \overline{B} + C ) (\overline{A} + \overline{\overline{B}} + \overline{\overline{C}}) \\
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A ( \overline{B} + C ) (\overline{A} + B + C) \\
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( A\overline{B} + AC ) (\overline{A} + B + C) \\
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A\overline{B}(\overline{A} + B + C) + AC(\overline{A} + B + C)\\
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(A\overline{B}\ \overline{A} + A\overline{B}B + A\overline{B}C) + (AC\overline{A} + ACB + ACC)\\
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(0 + 0 + A\overline{B}C) + (0 + ACB + AC)\\
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A\overline{B}C + ACB + AC\\
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ACB + AC\\
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AC
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\end{gather*}
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\exc{}
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LHS
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\begin{gather*}
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((A+B)+(A+C)) \overline{((A+B)(A+C))} \overline{A} \\
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(A+B+A+C) \overline{(A+B)(A+C)} \overline{A} \\
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(A+B+C) (\overline{(A+B)} + \overline{(A+C)}) \overline{A} \\
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(A+B+C) (\overline{A}\ \overline{B} + \overline{A}\ \overline{C}) \overline{A} \\
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(\overline{A}A + \overline{A}B + \overline{A}C) (\overline{A}\ \overline{B} + \overline{A}\ \overline{C}) \\
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(0 + \overline{A}B + \overline{A}C) (\overline{A}\ \overline{B} + \overline{A}\ \overline{C}) \\
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(\overline{A}B + \overline{A}C) (\overline{A}\ \overline{B} + \overline{A}\ \overline{C}) \\
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\overline{A}B \overline{A}\ \overline{B} +
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\overline{A}B \overline{A}\ \overline{C} +
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\overline{A}C \overline{A}\ \overline{B} +
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\overline{A}C \overline{A}\ \overline{C} \\
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0 +
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\overline{A}B \overline{A}\ \overline{C} +
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\overline{A}C \overline{A}\ \overline{B} +
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0 \\
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\overline{A}B \overline{C} + \overline{A}C \overline{B} \\
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\end{gather*}
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RHS
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\begin{gather*}
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(B+C) \overline{(BC)} \overline{A} \\
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(B+C) (\overline{B} + \overline{C}) \overline{A} \\
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(\overline{A}B + \overline{A}C) (\overline{B} + \overline{C}) \\
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\overline{A}B\overline{B} + \overline{A}B\overline{C} + \overline{A}C\overline{B} + \overline{A}C\overline{C} \\
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0 + \overline{A}B\overline{C} + \overline{A}C\overline{B} + 0 \\
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\overline{A}B\overline{C} + \overline{A}C\overline{B} \\
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\end{gather*}
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$LHS = RHS$
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\end{excs}
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\end{document}
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