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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} \definecolor{strokecol}{rgb}{0.0,0.0,0.0}; \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} \definecolor{strokecol}{rgb}{0.0,0.0,0.0}; \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} \definecolor{strokecol}{rgb}{0.0,0.0,0.0}; \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.
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