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37 lines
2.1 KiB
37 lines
2.1 KiB
We start by translating $\varphi$ to $\hat{\varphi} = \skel$ and assign the following variables to the theory literals:
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\begin{itemize}
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\item $e_{0}\Leftrightarrow(f(a)=f(b))$
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\item $e_{1}\Leftrightarrow(f(a)=c)$
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\item $e_{2}\Leftrightarrow(a=b)$
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\item $e_{3}\Leftrightarrow(f(c)=a)$
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\end{itemize}
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$\hat{\varphi} = (\clause{\lnot e_{0}; e_{1}; e_{2}}) \land (\clause{\lnot e_{1}; e_{2}; \lnot e_{3}}) \land (\clause{e_{0}; \lnot e_{1}}) \land (\clause{\lnot e_{0}; e_{2}}) \land (\clause{\lnot e_{2}; \lnot e_{3}}) \land (\clause{e_{0}; e_{2}}) \land (\clause{\lnot e_{2}; \lnot e_{1}}) $
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\hspace{-0.09cm}\scalebox{0.85}{
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\begin{dplltabular}{5}
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\dpllStep{1|2|3|4|5}
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\dpllDecL{0|0|1|1|1}
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\dpllAssi{ - |$\lnot e_{3}$|$\lnot e_{3}, \lnot e_{0}$|$\lnot e_{3}, \lnot e_{0}, \lnot e_{1}$|\makecell{$\lnot e_{3}, \lnot e_{0}, \lnot e_{1}, $ \\ $e_{2}$}}
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\dpllClause{1}{$\lnot e_{0}, e_{1}, e_{2}$}{$\lnot e_{0}, e_{1}, e_{2}$|$\lnot e_{0}, e_{1}, e_{2}$|\done|\done|\done}
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\dpllClause{2}{$\lnot e_{1}, e_{2}, \lnot e_{3}$}{$\lnot e_{1}, e_{2}, \lnot e_{3}$|\done|\done|\done|\done}
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\dpllClause{3}{$e_{0}, \lnot e_{1}$}{$e_{0}, \lnot e_{1}$|$e_{0}, \lnot e_{1}$|$\lnot e_{1}$|\done|\done}
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\dpllClause{4}{$\lnot e_{0}, e_{2}$}{$\lnot e_{0}, e_{2}$|$\lnot e_{0}, e_{2}$|\done|\done|\done}
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\dpllClause{5}{$\lnot e_{2}, \lnot e_{3}$}{$\lnot e_{2}, \lnot e_{3}$|\done|\done|\done|\done}
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\dpllClause{6}{$e_{0}, e_{2}$}{$e_{0}, e_{2}$|$e_{0}, e_{2}$|$e_{2}$|$e_{2}$|\done}
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\dpllClause{7}{$\lnot e_{2}, \lnot e_{1}$}{$\lnot e_{2}, \lnot e_{1}$|$\lnot e_{2}, \lnot e_{1}$|$\lnot e_{2}, \lnot e_{1}$|\done|\done}
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\dpllBCP{ - | - |$\lnot e_{1}$|$e_{2}$| - }
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\dpllPL{$\lnot e_{3}$| - | - | - | - }
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\dpllDeci{ - |$\lnot e_{0}$| - | - |SAT}
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\end{dplltabular}
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}
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$\Model_{\EUF} := (f(a) \neq f(b)) \land (f(a) \neq c) \land (a = b) \land (f(c) \neq a) $ \\
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Check if the assignment is consistent with the theory:
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\begin{align*}
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&\{a, b\}, \{f(a)\}, \{f(b)\}, \{c\}, \{f(c)\}
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\end{align*}
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$\Model_{\EUF}$ is consistent with the theory, \\$\Rightarrow \Model_{\EUF}$ is a satisfying assignment and $\varphi$ is SAT.
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