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Suppose that $A =\left\{1,2,3\right\}$ , $B=\left\{4,5,6\right\}$

$R= \left\{(1,4),(1,5),(2,5),(3,6)\right\}$ and $S= \left\{(4,5),(4,6),(5,4),(6,6)\right\}$

Notice that that $R$ is a relation from $A$ to $B$, and $S$ is a relation from $B$ to $B$.

Find $S\circ R$

I know the definition should be something like this

$S\circ R = \left\{(a,b) A \times B \mid \exists c \in C((a,c) \in R \land (c,b) \in S)\right\}$

The answer set in back of the book is

$\left\{(1,4),(1,5),(1,6),(2,4),(3,6)\right\}$

but I am having a hard time visualizing or figuring out how they got the answer since i am assuming $c$ must equal $b$ which is an element of $B$. I tried drawing the relations with pictures but still can't make sense of it. Any help in understanding how I should be thinking would come in handy.

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    $\begingroup$ In your case, $S\circ R = \left\{(a,b) A \times B \mid \exists c \in B,\,(a,c) \in R,\, (c,b) \in S\right\}$ hence if $a=1$ then $c$ can be ... hence ... $\endgroup$ – Did May 5 '17 at 5:48
  • $\begingroup$ Great, I see it now. Just matching the variables to the definition. $\endgroup$ – Squanchinator May 5 '17 at 6:30
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Draw $6$ nodes and label them $\{1,2,\ldots\}$. Represent $R$ by drawing an arrow joining node $x$ to $y$ for any $(x,y)$ in $R$. Do the same for $S$.

The composite relation $S\circ R$ is formed by all pairs $(x,z)$ where you start at $x$, follow an arrow to reach $y$ and then follow another arrow to reach $z$.

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  • $\begingroup$ FWIW, it would be more convenient to draw A and B and B in this order, to draw arrows from A to the first copy of B and from the first copy of B to the second copy of B, and then to follow the arrows. $\endgroup$ – Did May 5 '17 at 6:36
  • $\begingroup$ Not sure,which approach is better but they both makes sense. $\endgroup$ – Squanchinator May 5 '17 at 6:47

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