Take the 2-minute tour ×  This is from Velleman's "How to Prove
It", exercise 4.2.11.b). The exercise requires finding a mistake
in the proof, but everything looks good to me. Must be that I'm
missing some important fact, but any pointers are well
appreciated.  $R$ is relation from $A$ to $B$, and $S$ and $T$
are relations from $B$ to $C$.  Proof. Suppose $(a,c)\in
(S\setminus T)\circ R$. Then we can chose some $b\in B$ such
that $(a,b)\in R$ and $(b,c)\in S\setminus T$, so $(b,c)\in S$
and $(b,c)\not\in T$. Since $(a,b)\in R$ and $(b,c)\in S$,
$(a,c)\in S\circ R$. Similarly, since $(a,b)\in R$ and
$(b,c)\not\in T$, $(a,c)\not\in T\circ R$.Therefore $(a,c)\in
(S\circ R)\setminus(T\circ R)$. Since $(a,c)$ was arbitrary,
this shows that $(S\setminus T)\circ R\subseteq(S\circ
R)\setminus(T\circ R)$.  What am I missing here?  P.S. The
author does not imply that the proposition is true, so I cannot
say if there actually is a proof for this.  share|improve this
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