\hspace{-0.09cm}\scalebox{0.85}{
	\begin{dplltabular}{6}
		\dpllStep{7|8|9|10|11|12}
		\dpllDecL{0|1|1|1|1|1}
		\dpllAssi{ - |$\lnot e_{0}$|$\lnot e_{0}, e_{1}$|$\lnot e_{0}, e_{1}, e_{5}$|\makecell{$\lnot e_{0}, e_{1}, e_{5}, $ \\ $\lnot e_{4}$}|\makecell{$\lnot e_{0}, e_{1}, e_{5}, $ \\ $\lnot e_{4}, \lnot e_{2}$}}
		\dpllClause{1}{$\lnot e_{0}, \lnot e_{1}$}{$\lnot e_{0}, \lnot e_{1}$|\done|\done|\done|\done|\done}
		\dpllClause{2}{$e_{0}, e_{1}$}{$e_{0}, e_{1}$|$e_{1}$|\done|\done|\done|\done}
		\dpllClause{3}{$e_{2}, e_{3}, \lnot e_{4}$}{$e_{2}, e_{3}, \lnot e_{4}$|$e_{2}, e_{3}, \lnot e_{4}$|$e_{2}, e_{3}, \lnot e_{4}$|$e_{2}, e_{3}, \lnot e_{4}$|\done|\done}
		\dpllClause{4}{$e_{1}, \lnot e_{5}$}{$e_{1}, \lnot e_{5}$|$e_{1}, \lnot e_{5}$|\done|\done|\done|\done}
		\dpllClause{5}{$e_{3}, e_{5}, e_{4}$}{$e_{3}, e_{5}, e_{4}$|$e_{3}, e_{5}, e_{4}$|$e_{3}, e_{5}, e_{4}$|\done|\done|\done}
		\dpllClause{6}{$e_{0}, e_{5}$}{$e_{0}, e_{5}$|$e_{5}$|$e_{5}$|\done|\done|\done}
		\dpllClause{7}{$\lnot e_{2}, \lnot e_{3}, e_{0}, \lnot e_{1}, \lnot e_{5}$}{\makecell{$\lnot e_{2}, \lnot e_{3}, e_{0}, $ \\ $\lnot e_{1}, \lnot e_{5}$}|\makecell{$\lnot e_{2}, \lnot e_{3}, \lnot e_{1}, $ \\ $\lnot e_{5}$}|$\lnot e_{2}, \lnot e_{3}, \lnot e_{5}$|$\lnot e_{2}, \lnot e_{3}$|$\lnot e_{2}, \lnot e_{3}$|\done}
		\dpllBCP{ - |$e_{1}$|$e_{5}$| - | - | - }
		\dpllPL{ - | - | - |$\lnot e_{4}$|$\lnot e_{2}$| - }
		\dpllDeci{$\lnot e_{0}$| - | - | - | - |SAT}
	\end{dplltabular}
}

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

\begin{align*}
	&\{y, f(y)\}, \{z, x\}, \{f(x)\}
\end{align*}

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