We start by translating $\varphi$ to $\hat{\varphi} = \skel$ and assign the following variables to the theory literals:
\begin{itemize}
	\item $e_{0}\Leftrightarrow(f(a)=f(b))$
	\item $e_{1}\Leftrightarrow(f(a)=c)$
	\item $e_{2}\Leftrightarrow(a=b)$
	\item $e_{3}\Leftrightarrow(f(c)=a)$
\end{itemize}

$\hat{\varphi} = (\clause{\lnot e_{0}; e_{1}; e_{2}}) \land (\clause{\lnot e_{1}; e_{2}; \lnot e_{3}}) \land (\clause{e_{0}; \lnot e_{1}}) \land (\clause{\lnot e_{0}; e_{2}}) \land (\clause{\lnot e_{2}; \lnot e_{3}}) \land (\clause{e_{0}; e_{2}}) \land (\clause{\lnot e_{2}; \lnot e_{1}}) $


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

$\Model_{\EUF} := (f(a) \neq f(b)) \land (f(a) \neq c) \land (a = b) \land (f(c) \neq a) $ \\
Check if the assignment is consistent with the theory:

\begin{align*}
	&\{a, b\}, \{f(a)\}, \{f(b)\}, \{c\}, \{f(c)\}
\end{align*}

$\Model_{\EUF}$ is consistent with the theory, \\$\Rightarrow \Model_{\EUF}$ is a satisfying assignment and $\varphi$ is SAT.