assignments/week1/1.tex

@@ -1,6 +1,8 @@
 \documentclass[../main_text.tex]{subfiles}
 \begin{document}
 
+\section*{Week 1}
+
 \exercise*[0.3.6]
 \begin{enumerate}[label=\emph{\alph*)}]
     \item Wanting to show:
@@ -8,15 +10,40 @@
 
         In order to prove this equivalence, we have to prove the implication both ways. We use two lemmas for this.
 
-        \begin{lemma}[test]
+        \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}
-        \begin{definition}
-            Test.
-        \end{definition}
-    \item
+        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}