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sav08:extending_languages_of_decidable_theories [2008/04/15 14:27] vkuncak |
sav08:extending_languages_of_decidable_theories [2015/04/21 17:30] |
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- | ===== Extending Languages of Decidable Theories ===== | ||
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- | Recall from [[Quantifier elimination definition]] that if $T$ has effective quantifier elimination and there is an algorithm for deciding validity of ground formulas, then the theory is decidable. Now we show a sort of converse. | ||
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- | **Lemma:** Consider a set of formulas $T$. Then there exists an extended language $L' \supseteq L$ and a set of formulas $T'$ such that $T'$ has effective quantifier elimination, and such that $Conseq(T)$ is exactly the set of those formulas in $Conseq(T')$ that contain only symbols from ${\cal L}$. | ||
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- | **Proof:** | ||
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- | How to define $L'$? | ||
- | ++++| | ||
- | Fix an ordering on the set $V$ of all first-order variables. If $G$ is a term, formula, or a set of formulas, let $fv(G)$ denote the ordered list of its free variables. | ||
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- | Let ${\cal F}$ denote the set of formulas in language ${\cal L}$. Define | ||
- | \[ | ||
- | L' = \{ R_{F} \mid F \in {\cal F} \} | ||
- | \] | ||
- | where $R_{F}$ is a new relation symbol whose arity is equal to $|FV(F)|$. In other words, we introduce a relation symbol for each formula of the original language. | ||
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- | For each formula $F$ in language $L'$ we define relation-form of $F$, denoted $rf(F)$, which is a formula in the original language $L$ given by | ||
- | \[ | ||
- | \begin{array}{l} | ||
- | rf(F_1 \land F_2) = R_{rf(F_1) \land rf(F_2)} \\ | ||
- | rf(F_1 \lor F_2) = R_{rf(F_1) \lor rf(F_2)} \\ | ||
- | rf(\lnot F_1) = R_{\lnot rf(F_1)} \\ | ||
- | rf(\forall x. F_1) = R_{\forall x.rf(F_1)} \\ | ||
- | rf(\exists x. F_1) = R_{\exists x.rf(F_1)} \\ | ||
- | rf(\, R_{F(x_1,\ldots,x_n)}(t_1,\ldots,t_n) \,) = R_{F(t_1,\ldots,t_n)} | ||
- | \end{array} | ||
- | \] | ||
- | in the last definition, $F$ is a formula in $L$ and $fv(F) = x_1,\ldots,x_n$ is the sorted list of its free variables. | ||
- | ++++ | ||
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- | How to define $T'$? | ||
- | ++++| | ||
- | We let | ||
- | \[ | ||
- | T' = \{ F \mid rf(F) \in Conseq(T) \} | ||
- | \] | ||
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- | How to do quantifier elimination in $T'$? | ||
- | ++++| | ||
- | The quantifier-free version of $F$ is then simply $R_{rf(F)}(x_1,\ldots,x_n)$ where $fv(rf(F)) = x_1,\ldots,x_n$. This quantifier elimination is easy and effective (and trivial). | ||
- | ++++ | ||
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- | ++++ | ||
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- | **End Proof.** | ||
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- | Question: When is the question $F \in Conseq(T')$ for $T'$ from the proof decidable for ground formulas? ++|exactly when $F \in Conseq(T)$ is decidable.++ | ||
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- | Note: often we can have nicer representations instead of $R_F$, but this depends on particular theory. | ||