Questionnaire-2024/smt/3016_sol.tex

We start by translating $\varphi$ to $\hat{\varphi} = \skel$ and assign the following variables to the theory literals:
\begin{itemize}
	\item $e_{0}\Leftrightarrow(f(x)=x)$
	\item $e_{1}\Leftrightarrow(f(x)=y)$
	\item $e_{2}\Leftrightarrow(x=y)$
	\item $e_{3}\Leftrightarrow(z=f(x))$
\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_{1}}) \land (\clause{\lnot e_{3}; \lnot e_{1}}) $


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

Conflict in step 4\\
\scalebox{0.75}{

\begin{tikzpicture}[>=latex,line join=bevel,]
  \pgfsetlinewidth{1bp}
%%
\pgfsetcolor{black}
  % Edge: 1 -> 4
  \draw [->] (21.08bp,83.506bp) .. controls (26.44bp,80.758bp) and (33.44bp,77.508bp)  .. (40.0bp,75.5bp) .. controls (51.601bp,71.949bp) and (65.006bp,69.732bp)  .. (85.898bp,67.281bp);
  \definecolor{strokecol}{rgb}{0.0,0.0,0.0};
  \pgfsetstrokecolor{strokecol}
  \draw (54.0bp,83.0bp) node {$$1$$};
  % Edge: 1 -> 5
  \draw [->] (21.966bp,90.636bp) .. controls (35.227bp,93.477bp) and (58.98bp,98.567bp)  .. (86.047bp,104.37bp);
  \draw (54.0bp,107.0bp) node {$$5$$};
  % Edge: 2 -> 4
  \draw [->] (21.717bp,33.396bp) .. controls (33.05bp,36.904bp) and (52.243bp,43.253bp)  .. (68.0bp,50.5bp) .. controls (71.462bp,52.092bp) and (75.062bp,53.969bp)  .. (87.382bp,61.048bp);
  \draw (54.0bp,58.0bp) node {$$1$$};
  % Edge: 4 -> 3
  \draw [->] (107.67bp,69.723bp) .. controls (115.24bp,72.246bp) and (126.02bp,75.839bp)  .. (145.09bp,82.197bp);
  % Edge: 5 -> 3
  \draw [->] (107.67bp,102.94bp) .. controls (115.41bp,100.08bp) and (126.51bp,95.995bp)  .. (145.47bp,89.01bp);
  % Node: 1
\begin{scope}
  \definecolor{strokecol}{rgb}{0.0,0.0,0.0};
  \pgfsetstrokecolor{strokecol}
  \draw (11.0bp,88.5bp) ellipse (11.0bp and 11.0bp);
  \draw (11.0bp,88.5bp) node {$\lnot e_{1}$};
\end{scope}
  % Node: 4
\begin{scope}
  \definecolor{strokecol}{rgb}{0.0,0.0,0.0};
  \pgfsetstrokecolor{strokecol}
  \draw (97.0bp,66.5bp) ellipse (11.0bp and 11.0bp);
  \draw (97.0bp,66.5bp) node {$e_{2}$};
\end{scope}
  % Node: 5
\begin{scope}
  \definecolor{strokecol}{rgb}{0.0,0.0,0.0};
  \pgfsetstrokecolor{strokecol}
  \draw (97.0bp,106.5bp) ellipse (11.0bp and 11.0bp);
  \draw (97.0bp,106.5bp) node {$\lnot e_{2}$};
\end{scope}
  % Node: 2
\begin{scope}
  \definecolor{strokecol}{rgb}{0.0,0.0,0.0};
  \pgfsetstrokecolor{strokecol}
  \draw (11.0bp,30.5bp) ellipse (11.0bp and 11.0bp);
  \draw (11.0bp,30.5bp) node {$\lnot e_{0}$};
\end{scope}
  % Node: 3
\begin{scope}
  \definecolor{strokecol}{rgb}{0.0,0.0,0.0};
  \pgfsetstrokecolor{strokecol}
  \draw (156.0bp,85.5bp) ellipse (11.0bp and 11.0bp);
  \draw (156.0bp,85.5bp) node {$\bot$};
\end{scope}
%
\end{tikzpicture}


}
\begin{prooftree}
	\AxiomC{$1. \; e_{0} \lor e_{1} \lor e_{2}$}
	\AxiomC{$5. \; \lnot e_{2} \lor e_{1}$}
	\BinaryInfC{$e_{0} \lor e_{1}$}
\end{prooftree}

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

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

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

$\Model_{\EUF}$ is not consistent with the theory, because of: $(f(x) \neq x) $\\
$\Rightarrow$ We need to add a blocking clause from $\Model_{\EUF}: \\ $
$BC :=e_{0} \lor \neg e_{1} \lor \neg e_{2} \lor e_{3}$\\