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98 lines
3.7 KiB
98 lines
3.7 KiB
Conflict in step 8\\
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\scalebox{0.75}{
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\begin{tikzpicture}[>=latex,line join=bevel,]
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\pgfsetlinewidth{1bp}
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%%
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\pgfsetcolor{black}
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% Edge: 1 -> 4
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\draw [->] (22.3bp,31.0bp) .. controls (35.47bp,31.0bp) and (58.583bp,31.0bp) .. (85.878bp,31.0bp);
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\definecolor{strokecol}{rgb}{0.0,0.0,0.0};
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\pgfsetstrokecolor{strokecol}
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\draw (54.0bp,38.5bp) node {$$7$$};
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% Edge: 3 -> 2
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\draw [->] (193.67bp,47.438bp) .. controls (201.41bp,44.585bp) and (212.51bp,40.495bp) .. (231.47bp,33.51bp);
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% Edge: 4 -> 3
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\draw [->] (107.97bp,33.373bp) .. controls (121.23bp,36.53bp) and (144.98bp,42.186bp) .. (172.05bp,48.63bp);
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\draw (140.0bp,50.5bp) node {$$5$$};
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% Edge: 4 -> 5
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\draw [->] (107.11bp,26.371bp) .. controls (112.48bp,23.828bp) and (119.48bp,20.828bp) .. (126.0bp,19.0bp) .. controls (137.65bp,15.733bp) and (151.06bp,13.758bp) .. (171.92bp,11.639bp);
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\draw (140.0bp,26.5bp) node {$$8$$};
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% Edge: 5 -> 2
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\draw [->] (193.67bp,14.223bp) .. controls (201.24bp,16.746bp) and (212.02bp,20.339bp) .. (231.09bp,26.697bp);
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% Node: 1
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\begin{scope}
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\definecolor{strokecol}{rgb}{0.0,0.0,0.0};
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\pgfsetstrokecolor{strokecol}
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\draw (11.0bp,31.0bp) ellipse (11.0bp and 11.0bp);
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\draw (11.0bp,31.0bp) node {$\lnot e_{0}$};
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\end{scope}
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% Node: 4
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\begin{scope}
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\definecolor{strokecol}{rgb}{0.0,0.0,0.0};
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\pgfsetstrokecolor{strokecol}
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\draw (97.0bp,31.0bp) ellipse (11.0bp and 11.0bp);
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\draw (97.0bp,31.0bp) node {$e_{1}$};
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\end{scope}
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% Node: 2
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\begin{scope}
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\definecolor{strokecol}{rgb}{0.0,0.0,0.0};
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\pgfsetstrokecolor{strokecol}
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\draw (242.0bp,30.0bp) ellipse (11.0bp and 11.0bp);
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\draw (242.0bp,30.0bp) node {$\bot$};
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\end{scope}
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% Node: 3
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\begin{scope}
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\definecolor{strokecol}{rgb}{0.0,0.0,0.0};
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\pgfsetstrokecolor{strokecol}
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\draw (183.0bp,51.0bp) ellipse (11.0bp and 11.0bp);
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\draw (183.0bp,51.0bp) node {$e_{2}$};
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\end{scope}
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% Node: 5
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\begin{scope}
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\definecolor{strokecol}{rgb}{0.0,0.0,0.0};
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\pgfsetstrokecolor{strokecol}
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\draw (183.0bp,11.0bp) ellipse (11.0bp and 11.0bp);
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\draw (183.0bp,11.0bp) node {$\lnot e_{2}$};
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\end{scope}
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%
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\end{tikzpicture}
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}
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\begin{prooftree}
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\AxiomC{$5. \; \lnot e_{1} \lor e_{2}$}
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\AxiomC{$8. \; e_{0} \lor \lnot e_{1} \lor \lnot e_{2}$}
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\BinaryInfC{$\lnot e_{1} \lor e_{0}$}
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\AxiomC{$7. \; e_{0} \lor e_{1}$}
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\BinaryInfC{$e_{0}$}
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\end{prooftree}
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\hspace{-0.09cm}\scalebox{0.85}{
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\begin{dplltabular}{4}
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\dpllStep{9|10|11|12}
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\dpllDecL{0|0|0|0}
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\dpllAssi{ - |$e_{0}$|$e_{0}, e_{3}$|$e_{0}, e_{3}, e_{2}$}
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\dpllClause{1}{$e_{0}, e_{1}, \lnot e_{2}$}{$e_{0}, e_{1}, \lnot e_{2}$|\done|\done|\done}
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\dpllClause{2}{$e_{2}, e_{3}, e_{0}$}{$e_{2}, e_{3}, e_{0}$|\done|\done|\done}
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\dpllClause{3}{$\lnot e_{0}, e_{3}$}{$\lnot e_{0}, e_{3}$|$e_{3}$|\done|\done}
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\dpllClause{4}{$e_{2}, \lnot e_{3}, \lnot e_{0}$}{$e_{2}, \lnot e_{3}, \lnot e_{0}$|$e_{2}, \lnot e_{3}$|$e_{2}$|\done}
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\dpllClause{5}{$\lnot e_{1}, e_{2}$}{$\lnot e_{1}, e_{2}$|$\lnot e_{1}, e_{2}$|$\lnot e_{1}, e_{2}$|\done}
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\dpllClause{6}{$\lnot e_{1}, e_{2}, \lnot e_{3}$}{$\lnot e_{1}, e_{2}, \lnot e_{3}$|$\lnot e_{1}, e_{2}, \lnot e_{3}$|$\lnot e_{1}, e_{2}$|\done}
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\dpllClause{7}{$e_{0}, e_{1}$}{$e_{0}, e_{1}$|\done|\done|\done}
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\dpllClause{8}{$e_{0}, \lnot e_{1}, \lnot e_{2}$}{$e_{0}, \lnot e_{1}, \lnot e_{2}$|\done|\done|\done}
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\dpllClause{9}{$e_{0}$}{$e_{0}$|\done|\done|\done}
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\dpllBCP{$e_{0}$|$e_{3}$|$e_{2}$| - }
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\dpllPL{ - | - | - | - }
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\dpllDeci{ - | - | - |SAT}
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\end{dplltabular}
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}
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$\Model_{\EUF} := (f(a) = b) \land (b = c) \land (a = b) $ \\
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Check if the assignment is consistent with the theory:
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\begin{align*}
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&\{f(a), b\}, \{b, c\}, \{a, b\}\\
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&\{a, b, c, f(a)\}
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\end{align*}
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$\Model_{\EUF}$ is consistent with the theory, \\
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$\Rightarrow$ $\Model_{\EUF}$ is a satisfying assignment and $\varphi$ is SAT.
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