1.1 --- a/exercises/solutions/sol01.tex	Thu Apr 26 00:25:00 2012 +0200
     1.2 +++ b/exercises/solutions/sol01.tex	Fri Apr 27 02:02:08 2012 +0200
     1.3 @@ -8,10 +8,13 @@
     1.4  \usepackage{fullpage}
     1.5  \usepackage{mathtools}
     1.6  \usepackage[all]{xy}
     1.7 +\addtolength{\voffset}{-25pt}
     1.8  \title{CSP Exercise 01 Solution}
     1.9  \author{Eugen Sawin}
    1.10  \renewcommand{\familydefault}{\sfdefault}
    1.11  \newcommand{\E}{\mathcal{E}}
    1.12 +\newcommand{\R}{\mathcal{R}}
    1.13 +
    1.14  %\include{pythonlisting}
    1.15  
    1.16  \pagestyle{empty}
    1.17 @@ -42,10 +45,26 @@
    1.18  \end{itemize}
    1.19  
    1.20  \section*{Exercise 1.2}
    1.21 -(a) $R_{x,y} \bowtie S_{y,z} = \{(a,b,a),(a,b,c)\}$\\
    1.22 -(b) $\sigma_{z=c}(R_{x,y} \bowtie S_{y,z}) = \{(a,b,c)\}$\\
    1.23 -(c) $\pi_x(R_{x,y}) = \{(a)\}$\\
    1.24 -(d) $R_{x,y} \circ S_{y,z} = \{(b,b)\}$\\
    1.25 +(a) $R_{x,y} \bowtie S_{y,z} = \{(a,b,a),(a,b,c)\}$\\\\
    1.26 +(b) $\sigma_{z=c}(R_{x,y} \bowtie S_{y,z}) = \{(a,b,c)\}$\\\\
    1.27 +(c) $\pi_x(R_{x,y}) = \{(a)\}$\\\\
    1.28 +(d) $R_{x,y} \circ S_{y,z} = \{(a,a),(a,c)\}$\\\\
    1.29  
    1.30 -
    1.31 +\section*{Exercise 1.3}
    1.32 +(a) By definition holds
    1.33 +\begin{align}
    1.34 +  R \circ (S \cup T) &= \{(x,z) \in X^2 \mid \exists{y \in X}: (x,y) \in R \text{ and } (y,z) \in (S \cup T)\}\\
    1.35 +  R \circ S &= \{(x,z) \in X^2 \mid \exists{y \in X}: (x,y) \in R \text{ and } (y,z) \in S\}\\
    1.36 +  R \circ T &= \{(x,z) \in X^2 \mid \exists{y \in X}: (x,y) \in R \text{ and } (y,z) \in T\}
    1.37 +\end{align}
    1.38 +The union of (2) and (3) gives us the set of all $(x,y) \in R$ with either $(y,z) \in S$ or $(y,z) \in T$, from which follows that $(y,z) \in (S \cup T)$, making it equal to (1). \qed\\\\
    1.39 +(b) $-R$ was nowhere defined yet.\\\\
    1.40 +(c) By definition holds
    1.41 +\begin{align}
    1.42 +  (R \circ S)^{-1} &= \{(z,x) \in X^2 \mid \exists{y \in X}: (x,y) \in R \text{ and } (y,z) \in S\}\\
    1.43 +  S^{-1} \circ R^{-1} &= \{(x,z) \in X^2 \mid \exists{y \in X}: (x,y) \in S^{-1} \text{ and } (y,z) \in R^{-1}\}\\
    1.44 +  S^{-1} \circ R^{-1} &= \{(x,z) \in X^2 \mid \exists{y \in X}: (y,x) \in S \text{ and } (z,y) \in R\}
    1.45 +\end{align}
    1.46 +where (6) is (5) transformed by application of the definition of the converse of a relation. After some variable juggling we see that (4) and (6) are equal sets. \qed\\\\
    1.47 +(d) Todo 
    1.48  \end{document}