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