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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.
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