\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.