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remove trailing whitespace
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@ -1,4 +1,4 @@
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\documentclass{article}
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\documentclass{article}
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\usepackage{hyperref}
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\begin{document}
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@ -258,7 +258,7 @@ in Section \ref{follows}.
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\paragraph{Proof that $\mathbf{AXY=V \Rightarrow A'X'Y'=VD}$.}
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Substituting $A'X'Y'=(AB)(f(B,X))(f(f(X,B),Y))$, we
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recall that merges are commutative. So for any two
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changesets $P$ and $Q$,
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changesets $P$ and $Q$,
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$$m(P,Q)=m(Q,P)=Qf(Q,P)=Pf(P,Q)$$
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Applying this to the relation above, we see
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@ -267,7 +267,7 @@ A'X'Y'&=& AB f(B,X) f(f(X,B),Y) \\
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&=&AX f(X,B) f(f(X,B),Y) \\
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&=&A X Y f(Y, f(X,B)) \\
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&=&A X Y D \\
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&=&V D
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&=&V D
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\end{eqnarray*}
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As claimed.
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@ -348,7 +348,7 @@ the client's changeset $C$, it does five things:
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server's most recent revision number, which we call
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$r_H$ ($H$ for HEAD). $C'$ can be computed using
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follows (Section \ref{follows}). Remember that the server has a series of
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changesets,
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changesets,
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$$S_0\rightarrow S_1\rightarrow \ldots S_{r_c}\rightarrow S_{r_c+1} \rightarrow \ldots \rightarrow S_{r_H} $$
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$C$ is relative to $S_{r_c}$, but we need to compute $C'$ relative to $S_{r_H}$.
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We can compute a new $C$ relative to $S_{r_c+1}$ by computing $f(S_{r_c+1},C)$. Similarly we can repeat for
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