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sav08:interpretation_quotient_under_congruence [2008/04/02 22:20]
vkuncak
sav08:interpretation_quotient_under_congruence [2008/04/16 09:22]
maysam
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 ====== Quotient of an Interpretation under a Congruence ====== ====== Quotient of an Interpretation under a Congruence ======
 +
  
 ===== Example: quotient on pairs of natural numbers ===== ===== Example: quotient on pairs of natural numbers =====
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 This construction is an algebraic approach to construct from natural numbers one well-known structure. ​ Which one? ++| $({\cal Z}, + , -)$ where ${\cal Z}$ is the set of integers. ++ This construction is an algebraic approach to construct from natural numbers one well-known structure. ​ Which one? ++| $({\cal Z}, + , -)$ where ${\cal Z}$ is the set of integers. ++
  
-Note: this construction can be applied whenever we have an associative and commutative operation $*$ satisfying the cancelation law $x * z = y * z \rightarrow x=y$.  It allows us to contruct ​a structure where operation $*$ has an inverse. ​ What do we obtain if we apply this construction to multiplication of strictly positive integers?+Note: this construction can be applied whenever we have an associative and commutative operation $*$ satisfying the cancelation law $x * z = y * z \rightarrow x=y$.  It allows us to construct ​a structure where operation $*$ has an inverse. ​ What do we obtain if we apply this construction to multiplication of strictly positive integers?
  
 ===== Definition of Quotient of an Interpretation ===== ===== Definition of Quotient of an Interpretation =====
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 \] \]
  
-**Lemma 1:** $I_Q = ([D],I_Q)$ is a well-defined interpretation,​ that is, +**Lemma 1:** For each function symbol $f$ with $ar(f)=n$, the relation $\alpha_Q(f)$ is a total function $[D]^n \to [D]$ and for all $x_1,​\ldots,​x_n \in D$, 
-  * $[D]$ is non-empty;​ +\[ 
-  * for each function symbol $f$ with $ar(f)=n$, the relation $\alpha_Q(f)$ is a total function $[D]^n \to [D]$.+    \alpha_Q(f)([x_1],​\ldots,​[x_n]) = [\alpha(f)(x_1,​\ldots,​x_n)] 
 +\]
  
 **Lemma 2:** For each term $t$ we have $e_T(t)(I_Q) = [e_T(t)(I)]$. **Lemma 2:** For each term $t$ we have $e_T(t)(I_Q) = [e_T(t)(I)]$.