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\hspace{-0.09cm}\scalebox{0.85}{ \begin{dplltabular}{5} \dpllStep{7|8|9|10|11} \dpllDecL{0|1|1|1|1} \dpllAssi{ - |$\lnot e_{0}$|$\lnot e_{0}, e_{1}$|$\lnot e_{0}, e_{1}, e_{3}$|\makecell{$\lnot e_{0}, e_{1}, e_{3}, $ \\ $\lnot e_{2}$}} \dpllClause{1}{$e_{0}, e_{1}$}{$e_{0}, e_{1}$|$e_{1}$|\done|\done|\done} \dpllClause{2}{$e_{2}, e_{3}, e_{4}$}{$e_{2}, e_{3}, e_{4}$|$e_{2}, e_{3}, e_{4}$|$e_{2}, e_{3}, e_{4}$|\done|\done} \dpllClause{3}{$e_{5}, \lnot e_{2}, \lnot e_{4}$}{$e_{5}, \lnot e_{2}, \lnot e_{4}$|$e_{5}, \lnot e_{2}, \lnot e_{4}$|$e_{5}, \lnot e_{2}, \lnot e_{4}$|$e_{5}, \lnot e_{2}, \lnot e_{4}$|\done} \dpllClause{4}{$\lnot e_{0}, \lnot e_{1}$}{$\lnot e_{0}, \lnot e_{1}$|\done|\done|\done|\done} \dpllClause{5}{$e_{1}, \lnot e_{3}$}{$e_{1}, \lnot e_{3}$|$e_{1}, \lnot e_{3}$|\done|\done|\done} \dpllClause{6}{$\lnot e_{5}, \lnot e_{2}, e_{3}, e_{0}, \lnot e_{1}$}{\makecell{$\lnot e_{5}, \lnot e_{2}, e_{3}, $ \\ $e_{0}, \lnot e_{1}$}|\makecell{$\lnot e_{5}, \lnot e_{2}, e_{3}, $ \\ $\lnot e_{1}$}|$\lnot e_{5}, \lnot e_{2}, e_{3}$|\done|\done} \dpllBCP{ - |$e_{1}$| - | - | - } \dpllPL{ - | - |$e_{3}$|$\lnot e_{2}$| - } \dpllDeci{$\lnot e_{0}$| - | - | - |SAT} \end{dplltabular}}
$\Model_{\EUF} := (a \neq b) \land (a = f(a)) \land (f(a) \neq b) \land (c = b) $ \\Check if the assignment is consistent with the theory:
\begin{align*} &\{a, f(a)\}, \{c, b\}\end{align*}
$\Model_{\EUF}$ is consistent with the theory, \\$\Rightarrow \Model_{\EUF}$ is a satisfying assignment and $\varphi$ is SAT.
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