\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}