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sav08:definition_of_presburger_arithmetic [2008/04/15 20:15]
vkuncak
sav08:definition_of_presburger_arithmetic [2015/04/21 17:30] (current)
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This theory is very simple to describe, but it is very far from allowing quantifier-elimination. ​ For example, it does not have a name for zero, so we cannot express $\forall x. x+y=x$. ​ It also does not have a way to express $\exists y. x+y=z$. This theory is very simple to describe, but it is very far from allowing quantifier-elimination. ​ For example, it does not have a name for zero, so we cannot express $\forall x. x+y=x$. ​ It also does not have a way to express $\exists y. x+y=z$.
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===== Language of Presburger Arithmetic that Admits QE ===== ===== Language of Presburger Arithmetic that Admits QE =====
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* introduce not only addition but also subtraction ​   * introduce not only addition but also subtraction ​
* to conveniently express certain expressions,​ introduce function $m_K$ for each $K \in {\cal Z}$, to be interpreted as multiplication by a constant, $m_K(x)= K \cdot x$.  We write $m_K$ as $K \cdot x$   * to conveniently express certain expressions,​ introduce function $m_K$ for each $K \in {\cal Z}$, to be interpreted as multiplication by a constant, $m_K(x)= K \cdot x$.  We write $m_K$ as $K \cdot x$
-  * to enable quantifier elimination from $\exists x. y= K \cdot x$ introduce for each $K$ predicate $d_K(y)$ (divisibility by constant)+  * to enable quantifier elimination from $\exists x. y= K \cdot x$ introduce for each $K$ predicate $K|y$ (divisibility by constant)

The resulting language The resulting language
-$+\begin{equation*} L = \{ +, -, < \} \cup \{ K \mid K \in \mathbb{Z}\} \cup \{ (K \cdot \_) \mid K \in \mathbb{Z} \} \cup \{ (K|\_) \mid K \in \mathbb{Z} \} L = \{ +, -, < \} \cup \{ K \mid K \in \mathbb{Z}\} \cup \{ (K \cdot \_) \mid K \in \mathbb{Z} \} \cup \{ (K|\_) \mid K \in \mathbb{Z} \} -$+\end{equation*}