We start by translating $\varphi$ to $\hat{\varphi} = \skel$ and assign the following variables to the theory literals:
\begin{itemize}
	\item $e_{0}\Leftrightarrow(a=x)$
	\item $e_{1}\Leftrightarrow(a=y)$
	\item $e_{2}\Leftrightarrow(x=y)$
	\item $e_{3}\Leftrightarrow(z=a)$
	\item $e_{4}\Leftrightarrow(b=z)$
\end{itemize}

$\hat{\varphi} = (\clause{e_{0}; e_{1}; e_{2}})\land (\clause{\lnot e_{0}; e_{1}})\land (\clause{\lnot e_{1}; e_{2}})\land (\clause{e_{2}; e_{3}})\land (\clause{\lnot e_{2}; e_{4}})\land (\clause{\lnot e_{3}; \lnot e_{4}})$

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

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

\begin{align*}
	&\{x, y\}, \{b, z\}, \{a\}
\end{align*}

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