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3 lines
1.2 KiB
3 lines
1.2 KiB
\item \self In the following text fill the blanks with the missing word(s).
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One way to solve \textit{Satisfiability} \rule{3cm}{0.15mm} \textit{Theories} problems works as follows. First, the propositional skeleton of the formula in question is given to a \rule{1.5cm}{0.15mm} solver. If this solver returns \rule{2.5cm}{0.15mm}, we terminate with answer \rule{2.5cm}{0.15mm}. In the other case, the solver returns a \rule{5cm}{0.15mm}, which is a \rule{3.5cm}{0.15mm} of theory literals. This can be given to a \rule{3cm}{0.15mm} solver that can decide the \rule{3cm}{0.15mm} fragment of the theory in question. If this solver returns \rule{2.5cm}{0.15mm}, we terminate with answer \rule{3.5cm}{0.15mm}. Otherwise, we add a \rule{2.5cm}{0.15mm} \rule{2.5cm}{0.15mm} to the propositional skeleton, to prevent the same \rule{4cm}{0.15mm} from occurring again, and run the \rule{1.5cm}{0.15mm} solver on the augmented propositional skeleton. This loop is repeated until either the \rule{1.5cm}{0.15mm} solver returns \rule{3.5cm}{0.15mm} (in which case the answer is \rule{3.5cm}{0.15mm}), or the \rule{3cm}{0.15mm} solver returns \rule{3.5cm}{0.15mm} (in which case the answer is \rule{3cm}{0.15mm}). This entire procedure is called \rule{2cm}{0.15mm} encoding.
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