1.1 --- a/exercises/solutions/sol01.tex	Fri Apr 27 02:02:08 2012 +0200
     1.2 +++ b/exercises/solutions/sol01.tex	Fri Apr 27 19:03:09 2012 +0200
     1.3 @@ -8,7 +8,7 @@
     1.4  \usepackage{fullpage}
     1.5  \usepackage{mathtools}
     1.6  \usepackage[all]{xy}
     1.7 -\addtolength{\voffset}{-25pt}
     1.8 +\addtolength{\voffset}{-40pt}
     1.9  \title{CSP Exercise 01 Solution}
    1.10  \author{Eugen Sawin}
    1.11  \renewcommand{\familydefault}{\sfdefault}
    1.12 @@ -48,7 +48,7 @@
    1.13  (a) $R_{x,y} \bowtie S_{y,z} = \{(a,b,a),(a,b,c)\}$\\\\
    1.14  (b) $\sigma_{z=c}(R_{x,y} \bowtie S_{y,z}) = \{(a,b,c)\}$\\\\
    1.15  (c) $\pi_x(R_{x,y}) = \{(a)\}$\\\\
    1.16 -(d) $R_{x,y} \circ S_{y,z} = \{(a,a),(a,c)\}$\\\\
    1.17 +(d) $R_{x,y} \circ S_{y,z} = \{(a,a),(a,c)\}$
    1.18  
    1.19  \section*{Exercise 1.3}
    1.20  (a) By definition holds
    1.21 @@ -56,15 +56,34 @@
    1.22    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)\}\\
    1.23    R \circ S &= \{(x,z) \in X^2 \mid \exists{y \in X}: (x,y) \in R \text{ and } (y,z) \in S\}\\
    1.24    R \circ T &= \{(x,z) \in X^2 \mid \exists{y \in X}: (x,y) \in R \text{ and } (y,z) \in T\}
    1.25 -\end{align}
    1.26 +\end{align}\setcounter{equation}{0}\\
    1.27  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\\\\
    1.28 -(b) $-R$ was nowhere defined yet.\\\\
    1.29 +(b) By definition holds
    1.30 +\begin{align}
    1.31 +  -R &= \{(x,y) \in X^2 \mid (x,y) \notin R\}\\
    1.32 +  (-R)^{-1} &= \{(y,x) \in X^2 \mid (x,y) \in -R\}\\
    1.33 +  -(R^{-1}) &= \{(x,y) \in X^2 \mid (x,y) \notin R^{-1}\}\\
    1.34 +  -(R^{-1}) &= \{(x,y) \in X^2 \mid (y,x) \notin R\}
    1.35 +\end{align}\setcounter{equation}{0}\\
    1.36 +We've obtained (4) by applying the converse definition on (3). From the definition of $-R$ (1) follows that $(x,y) \in -R$ iff $(x,y) \notin R$, and therefore we see that (2) and (4) define equal sets. \qed\\\\
    1.37  (c) By definition holds
    1.38  \begin{align}
    1.39    (R \circ S)^{-1} &= \{(z,x) \in X^2 \mid \exists{y \in X}: (x,y) \in R \text{ and } (y,z) \in S\}\\
    1.40    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}\}\\
    1.41    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\}
    1.42 -\end{align}
    1.43 -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\\\\
    1.44 -(d) Todo 
    1.45 +\end{align}\setcounter{equation}{0}\\
    1.46 +We've obtained (3) by applying the converse definition on (2). After some variable juggling we see that (1) and (3) define equal sets. \qed\\\\
    1.47 +(d) By definition holds
    1.48 +\begin{align}
    1.49 +  R \circ S &= \{(x,z) \in X^2 \mid \exists{y \in X}: (x,y) \in R \text{ and } (y,z) \in S\}\\
    1.50 +  (R \circ S) \cap T^{-1} &= (R \circ S) \cap \{(y,x) \in X^2 \mid (x,y) \in T\}\\
    1.51 +  (R \circ S) \cap T^{-1} &= \{(x,z) \in X^2 \mid \exists{y \in X}: (x,y) \in R \text{ and } (y,z) \in S \text{ and } (z,x) \in T\}\\
    1.52 +  (S \circ T) \cap R^{-1} &= \{(x,z) \in X^2 \mid \exists{y \in X}: (x,y) \in S \text{ and } (y,z) \in T \text{ and } (z,x) \in R\}
    1.53 +\end{align}\\
    1.54 +We've obtained (3) by directly applying intersection within the set comprehension of (2). It follows that\\
    1.55 +\begin{align}
    1.56 +  (R \circ S) \cap T^{-1} = \emptyset &\iff \forall{x,y,z \in X}: (x,y) \notin R \text{ or } (y,z) \notin S \text{ or } (z,x) \notin T\\
    1.57 +  \text{ and } (S \circ T) \cap R^{-1} = \emptyset &\iff \forall{x,y,z \in X}: (x,y) \notin S \text{ or } (y,z) \notin T \text{ or } (z,x) \notin R
    1.58 +\end{align}\\
    1.59 +As we can see, the intersections form a ring-like relationship between the sets' tuples $(...,R, S, T, R,...)$. After some variable reordering, we see that both intersections are empty iff there is no ring-like relationship between any tuples of the three sets. \qed
    1.60  \end{document}