Conflict in step 8\\
\scalebox{0.75}{

\begin{tikzpicture}[>=latex,line join=bevel,]
  \pgfsetlinewidth{1bp}
%%
\pgfsetcolor{black}
  % Edge: 1 -> 4
  \draw [->] (22.3bp,31.0bp) .. controls (35.47bp,31.0bp) and (58.583bp,31.0bp)  .. (85.878bp,31.0bp);
  \definecolor{strokecol}{rgb}{0.0,0.0,0.0};
  \pgfsetstrokecolor{strokecol}
  \draw (54.0bp,38.5bp) node {$$7$$};
  % Edge: 3 -> 2
  \draw [->] (193.67bp,47.438bp) .. controls (201.41bp,44.585bp) and (212.51bp,40.495bp)  .. (231.47bp,33.51bp);
  % Edge: 4 -> 3
  \draw [->] (107.97bp,33.373bp) .. controls (121.23bp,36.53bp) and (144.98bp,42.186bp)  .. (172.05bp,48.63bp);
  \draw (140.0bp,50.5bp) node {$$5$$};
  % Edge: 4 -> 5
  \draw [->] (107.11bp,26.371bp) .. controls (112.48bp,23.828bp) and (119.48bp,20.828bp)  .. (126.0bp,19.0bp) .. controls (137.65bp,15.733bp) and (151.06bp,13.758bp)  .. (171.92bp,11.639bp);
  \draw (140.0bp,26.5bp) node {$$8$$};
  % Edge: 5 -> 2
  \draw [->] (193.67bp,14.223bp) .. controls (201.24bp,16.746bp) and (212.02bp,20.339bp)  .. (231.09bp,26.697bp);
  % Node: 1
\begin{scope}
  \definecolor{strokecol}{rgb}{0.0,0.0,0.0};
  \pgfsetstrokecolor{strokecol}
  \draw (11.0bp,31.0bp) ellipse (11.0bp and 11.0bp);
  \draw (11.0bp,31.0bp) node {$\lnot e_{0}$};
\end{scope}
  % Node: 4
\begin{scope}
  \definecolor{strokecol}{rgb}{0.0,0.0,0.0};
  \pgfsetstrokecolor{strokecol}
  \draw (97.0bp,31.0bp) ellipse (11.0bp and 11.0bp);
  \draw (97.0bp,31.0bp) node {$e_{1}$};
\end{scope}
  % Node: 2
\begin{scope}
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  \pgfsetstrokecolor{strokecol}
  \draw (242.0bp,30.0bp) ellipse (11.0bp and 11.0bp);
  \draw (242.0bp,30.0bp) node {$\bot$};
\end{scope}
  % Node: 3
\begin{scope}
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  \pgfsetstrokecolor{strokecol}
  \draw (183.0bp,51.0bp) ellipse (11.0bp and 11.0bp);
  \draw (183.0bp,51.0bp) node {$e_{2}$};
\end{scope}
  % Node: 5
\begin{scope}
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  \pgfsetstrokecolor{strokecol}
  \draw (183.0bp,11.0bp) ellipse (11.0bp and 11.0bp);
  \draw (183.0bp,11.0bp) node {$\lnot e_{2}$};
\end{scope}
%
\end{tikzpicture}


}
\begin{prooftree}
	\AxiomC{$5. \; \lnot e_{1} \lor e_{2}$}
	\AxiomC{$8. \; e_{0} \lor \lnot e_{1} \lor \lnot e_{2}$}
	\BinaryInfC{$\lnot e_{1} \lor e_{0}$}
	\AxiomC{$7. \; e_{0} \lor e_{1}$}
	\BinaryInfC{$e_{0}$}
\end{prooftree}

\hspace{-0.09cm}\scalebox{0.85}{
	\begin{dplltabular}{4}
		\dpllStep{9|10|11|12}
		\dpllDecL{0|0|0|0}
		\dpllAssi{ - |$e_{0}$|$e_{0}, e_{3}$|$e_{0}, e_{3}, e_{2}$}
		\dpllClause{1}{$e_{0}, e_{1}, \lnot e_{2}$}{$e_{0}, e_{1}, \lnot e_{2}$|\done|\done|\done}
		\dpllClause{2}{$e_{2}, e_{3}, e_{0}$}{$e_{2}, e_{3}, e_{0}$|\done|\done|\done}
		\dpllClause{3}{$\lnot e_{0}, e_{3}$}{$\lnot e_{0}, e_{3}$|$e_{3}$|\done|\done}
		\dpllClause{4}{$e_{2}, \lnot e_{3}, \lnot e_{0}$}{$e_{2}, \lnot e_{3}, \lnot e_{0}$|$e_{2}, \lnot e_{3}$|$e_{2}$|\done}
		\dpllClause{5}{$\lnot e_{1}, e_{2}$}{$\lnot e_{1}, e_{2}$|$\lnot e_{1}, e_{2}$|$\lnot e_{1}, e_{2}$|\done}
		\dpllClause{6}{$\lnot e_{1}, e_{2}, \lnot e_{3}$}{$\lnot e_{1}, e_{2}, \lnot e_{3}$|$\lnot e_{1}, e_{2}, \lnot e_{3}$|$\lnot e_{1}, e_{2}$|\done}
		\dpllClause{7}{$e_{0}, e_{1}$}{$e_{0}, e_{1}$|\done|\done|\done}
		\dpllClause{8}{$e_{0}, \lnot e_{1}, \lnot e_{2}$}{$e_{0}, \lnot e_{1}, \lnot e_{2}$|\done|\done|\done}
		\dpllClause{9}{$e_{0}$}{$e_{0}$|\done|\done|\done}
		\dpllBCP{$e_{0}$|$e_{3}$|$e_{2}$| - }
		\dpllPL{ - | - | - | - }
		\dpllDeci{ - | - | - |SAT}
	\end{dplltabular}
}

$\Model_{\EUF} := (f(a) = b) \land (b = c) \land (a = b) $ \\
Check if the assignment is consistent with the theory:
\begin{align*}
	&\{f(a), b\}, \{b, c\}, \{a, b\}\\
	&\{a, b, c, f(a)\}
\end{align*}
$\Model_{\EUF}$ is consistent with the theory, \\
$\Rightarrow$ $\Model_{\EUF}$ is a satisfying assignment and $\varphi$ is SAT.