\documentclass[../main_text.tex]{subfiles} \begin{document} \section*{Week 1} \exercise*[0.3.6] \begin{enumerate}[label=\emph{\alph*)}] \item Wanting to show: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ In order to prove this equivalence, we have to prove the implication both ways. We use two lemmas for this. \begin{lemma} \label{lem:set1} $A \cap (B \cup C) \implies (A \cap B) \cup (A \cap C)$ \end{lemma} Let $x \in A \cap (B \cup C)$. By the definition of set intersection, $x \in A$ and $x \in B \cup C$. By the definition of set union, $x \in A$ and $(x \in B$ or $x \in C)$. From propositional logic we know that for propositions $P$, $Q$ and $R$ the following holds: $P \land (Q \lor R) \iff (P \land Q) \lor (P \land R)$. So, substituting for this particular case yields $(x \in A$ and $x \in B)$ or $(x \in A$ and $x \in C)$. Using the definition of set intersection again gets $x \in A \cap B$ or $x \in A \cap C$. Using the definition of set union again gives $x \in (A \cap B) \cup (A \cap C)$. \qed \begin{lemma} \label{lem:set2} $(A \cap B) \cup (A \cap C) \implies A \cap (B \cup C)$ \end{lemma} Let $x \in (A \cap B) \cup (A \cap C)$. By the definition of set union, $x \in (A \cap B)$ or $x \in (A \cap C)$. By the definition of set intersection, $(x \in A$ or $x \in B)$ and $(x \in A$ or $x \in B)$. Using the same propositional logical equivalence as in Lemma \ref{lem:set1}, this gives $x \in A$ and $(x \in B$ or $x \in C)$. Wrapping up, we use the definition of set union to get $x \in A$ and $x \in B \cup C$ and the definition of intersection to get $x \in A \cap (B \cup C)$. \qed Using Lemma \ref{lem:set1} and \ref{lem:set2}, we get the desired equivalence of $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$. \qed \item Wanting to show: $A \cup (B \cap C) = (A \cup B) \cap (a \cup C)$ This proof is so similar to a) that it feels like a waste of time and will therefore be left to the reader. \end{enumerate} \hrule \end{document}