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-\documentclass[12pt]{article}
-\usepackage{ntnu}
-\usepackage{ntnu-math}
-
-\author{Øystein Tveit}
-\title{MA0301 Exercise 2}
-
-\begin{document}
-  \ntnuTitle{}
-  \break{}
-  
-  \begin{excs}
-    \exc{}
-
-    \exc{}
- 
-    \exc{}
-    \begin{subexcs}
-      \subexc{}
-
-      \begin{ssubexcs}
-        \ssubexc{}
-        \begin{align*}
-          {{2,3,5} \cup {6,4}} &\cap {4,6,8} \\
-          {{2,4,6}} &\cap {4,6,8} \\
-          \emptyset
-        \end{align*}
-
-        \ssubexc{}
-        \begin{align*}
-          P({7,8,9}) &- P({7,9}) \\
-          {{7,8,9}, {7,8}, {8,9}, {7,9}, {7}, {8}, {9}, \emptyset} &- {{7,9}, {7}, {9}, \emptyset} \\
-          {{7,8,9}, {7,8}, {8,9}, {8}}
-        \end{align*}
-
-        \ssubexc{}
-        \begin{align*}
-          P(\emptyset) \\
-          {\emptyset}
-        \end{align*}
-
-        \ssubexc{}
-        \begin{align*}
-          {1, 3, 5} \times {0} \\
-          {	\langle 1,0	\rangle, \langle 3,0 \rangle, \langle 5, 0 \rangle }
-        \end{align*}
-        
-        \ssubexc{}
-        \begin{align*}
-          {2,4,6} \times \emptyset \\
-          \emptyset
-        \end{align*}
-
-        \ssubexc{}
-        \begin{align*}
-          P({0}) &\times P({1}) \\
-          {\emptyset, {0}} &\times {\emptyset, {1}} \\
-          {\langle\emptyset,\emptyset\rangle, \langle\emptyset,{1}\rangle, \langle{0},\emptyset\rangle, \langle{0},{1}\rangle}
-        \end{align*}
-        
-        \ssubexc{}
-        \begin{align*}
-          P(P({2})) \\
-          P({\emptyset,{2}}) \\
-          { {{\emptyset}, {2}}, {{\emptyset}}, {{2}}, \emptyset }
-        \end{align*}
-        
-      \end{ssubexcs}
-
-      \subexc{}
-      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
-      \[ |P(A) - {{x} : x \in A}| = \frac{n}{2} \]
-      
-    \end{subexcs}
-
-    \exc{}
-    \begin{subexcs}
-      \subexc{}
-      $\emptyset = {\emptyset}$ is {\color{red}False} because $|\emptyset| \neq |{\emptyset}|$
-
-      \subexc{}
-      $\emptyset = {0}$ is {\color{red}False} because $|\emptyset| \neq |{0}|$
-      
-      \subexc{}
-      $|\emptyset| = 0$ is {\color{ForestGreen}True} because $\emptyset$ has $0$ elements
-
-      \subexc{}
-      $P(\emptyset)$ is {\color{red}False} because $P(\emptyset) = {{\emptyset}}$ has $1$ element
-
-      \subexc{}
-      $\emptyset = {}$ is {\color{ForestGreen}True} because the empty set is a subset of every possible set
-
-      \subexc{}
-      $\emptyset = {x \in \mathbb{N} : x \leq 0 and x > 0}$ 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
-    \end{subexcs} 
-
-    \exc{}
-    \begin{subexcs}
-      \subexc{}
-      \begin{align*}
-        A \cap (\A \cup B) \\
-        {x : x \in A \wedge x \in (A \cup B)} \\
-        {x : x \in A \wedge (x \in A \or x \in B)} \\
-        {x : x \in A} \\
-        A
-      \end{align*}
-
-      \subexc{}
-      \begin{align*}
-        A-(B \cap C) \\
-        {x : x \in A \wedge x \notin (B \cap C)} \\
-        {x : x \in A \wedge (x \notin B \wedge x \notin C)} \\
-        {x : (x \in A \wedge x \notin B) \vee (x \in A \wedge x \notin C)} \\
-        {x : x \in (A - B) \vee x \in (A - C)} \\
-        {x : x \in (A - B) \cup (A - C)} \\
-        (A-B) \cup (A-C)
-      \end{align*}
-
-      \end{subexcs} 
-
-      \exc{}
-      \begin{subexcs}
-        \subexc{}
-        
-      \end{subexcs} 
-
-      \exc{}
-      \begin{align*}
-        X &= {{1,2,3}, {2,3}, {ef}} \cup {{e}} \\
-          &= {{1,2,3}, {2,3}, {ef}, {e}} \\
-          \\
-        P(x) &= {
-               {{1,2,3}, {2,3}, {ef}, {e}},
-               {{1,2,3}, {2,3}, {ef}},
-               {{1,2,3}, {2,3}, {e}},
-               {{1,2,3}, {ef}, {e}},
-               {{2,3}, {ef}, {e}},
-               {{1,2,3}, {2,3}}
-               {{1,2,3}, {e}}
-               {{1,2,3}, {ef}}
-               {{2,3}, {ef}}
-               {{2,3}, {e}}
-               {{ef}, {e}}
-               {{e}}
-               {{ef}},
-               {{2,3}},
-               {{1,2,3}}
-               } \\
-        \\
-       P(X \cap Y) &= P({{1,2,3}, {2,3}, {ef}, {e}} \cap {{1,2,3,e,f}}) \\
-                &= P(\emptyset) \\
-                &= {\emptyset}
-      \end{align*}      
-
-      \exc{}
-      \begin{subexcs}
-        \subexc{}
-        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.
-
-        \begin{align*}
-          A_1 \cap A_2 \cap A_3 \\
-          A_1 \cap A_2 \cap \overline{A_3} \\
-          A_1 \cap \overline{A_2} \cap A_3 \\
-          A_1 \cap \overline{A_2} \cap \overline{A_3} \\
-          \overline{A_1} \cap A_2 \cap A_3 \\
-          \overline{A_1} \cap A_2 \cap \overline{A_3} \\
-          \overline{A_1} \cap \overline{A_2} \cap A_3 \\
-          \overline{A_1} \cap \overline{A_2} \cap \overline{A_3} \\
-        \end{align*}
-
-      \subexc{}
-        For each set, the amount of fundamental products is multiplied by $2$. Therefore, the amount of fundamental sets of $n$ sets is $2^n$
-
-      \end{subexcs}
-
-      \exc{}
-      \begin{align*}
-        A\overline{(B\overline{C})}\overline{((A\overline{B})\overline{C})} \\
-        A\overline{(B\overline{C})}\overline{((A\overline{B})\overline{C})} \\
-      \end{align*}
-
-  \end{excs}
-  
-
-\end{document}