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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 $e_{0}\Leftrightarrow(y=f(x))$ \item $e_{1}\Leftrightarrow(y=f(y))$ \item $e_{2}\Leftrightarrow(f(x)=z)$ \item $e_{3}\Leftrightarrow(f(y)=x)$ \item $e_{4}\Leftrightarrow(f(y)=z)$ \item $e_{5}\Leftrightarrow(z=x)$\end{itemize}
$\hat{\varphi} = (\clause{\lnot e_{0}; \lnot e_{1}}) \land (\clause{e_{0}; e_{1}}) \land (\clause{e_{2}; e_{3}; \lnot e_{4}}) \land (\clause{e_{1}; \lnot e_{5}}) \land (\clause{e_{3}; e_{5}; e_{4}}) \land (\clause{e_{0}; e_{5}})$
\hspace{-0.09cm}\scalebox{0.85}{ \begin{dplltabular}{6} \dpllStep{1|2|3|4|5|6} \dpllDecL{0|0|0|1|1|1} \dpllAssi{ - |$e_{2}$|$e_{2}, e_{3}$|$e_{2}, e_{3}, \lnot e_{0}$|\makecell{$e_{2}, e_{3}, \lnot e_{0}, $ \\ $e_{1}$}|\makecell{$e_{2}, e_{3}, \lnot e_{0}, $ \\ $e_{1}, e_{5}$}} \dpllClause{1}{$\lnot e_{0}, \lnot e_{1}$}{$\lnot e_{0}, \lnot e_{1}$|$\lnot e_{0}, \lnot e_{1}$|$\lnot e_{0}, \lnot e_{1}$|\done|\done|\done} \dpllClause{2}{$e_{0}, e_{1}$}{$e_{0}, e_{1}$|$e_{0}, e_{1}$|$e_{0}, e_{1}$|$e_{1}$|\done|\done} \dpllClause{3}{$e_{2}, e_{3}, \lnot e_{4}$}{$e_{2}, e_{3}, \lnot e_{4}$|\done|\done|\done|\done|\done} \dpllClause{4}{$e_{1}, \lnot e_{5}$}{$e_{1}, \lnot e_{5}$|$e_{1}, \lnot e_{5}$|$e_{1}, \lnot e_{5}$|$e_{1}, \lnot e_{5}$|\done|\done} \dpllClause{5}{$e_{3}, e_{5}, e_{4}$}{$e_{3}, e_{5}, e_{4}$|$e_{3}, e_{5}, e_{4}$|\done|\done|\done|\done} \dpllClause{6}{$e_{0}, e_{5}$}{$e_{0}, e_{5}$|$e_{0}, e_{5}$|$e_{0}, e_{5}$|$e_{5}$|$e_{5}$|\done} \dpllBCP{ - | - | - |$e_{1}$|$e_{5}$| - } \dpllPL{$e_{2}$|$e_{3}$| - | - | - | - } \dpllDeci{ - | - |$\lnot e_{0}$| - | - | - } \end{dplltabular}}
$\Model_{\EUF} := (y \neq f(x)) \land (y = f(y)) \land (f(x) = z) \land (f(y) = x) \land (z = x) $ \\Check if the assignment is consistent with the theory:
\begin{align*} &\{y, f(y)\}, \{f(x), z\}, \{f(y), x\}, \{z, x\}\\ &\{f(x), f(y), x, y, z\}\end{align*}
$\Model_{\EUF}$ is not consistent with the theory, because of: $(y \neq f(x)) $\\$\Rightarrow$ We need to add a blocking clause from $\Model_{\EUF}: \\ $$BC :=\neg e_{2} \lor \neg e_{3} \lor e_{0} \lor \neg e_{1} \lor \neg e_{5}$\\
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