Quotient of an Interpretation under a Congruence

Example: quotient on pairs of natural numbers

Let ${\cal N} = \{0,1,2,\ldots, \}$ . Consider a structure with domain $N^2$ , with functions (plus and minus):

$\begin{equation*} p((x_1,y_1),(x_2,y_2)) = (x_1 + x_2, y_1 + y_2) \end{equation*}$

$\begin{equation*} m((x_1,y_1),(x_2,y_2)) = (x_1 + y_2, y_1 + x_2) \end{equation*}$

Relation $r$ defined by

$\begin{equation*} r = \{((x_1,y_1),(x_2,y_2)) \mid x_1 + y_2 = x_2 + y_1 \} \end{equation*}$

is a congruence with respect to operations $p$ and $m$ . Indded, we can check that, for example, if $r((x_1,y_1),(x'_1,y'_1))$ and $r((x_2,y_2),(x'_2,y'_2))$ then

$\begin{equation*} r(p((x_1,y_1),(x_2,y_2)), p((x'_1,y'_1),(x'_2,y'_2))) \end{equation*}$

Congruence is an equivalence relation. What are equivalence classes for elements:

$[(1,1)] =$ $\{ (x,y) \mid x=y \}$

$[(10,1)] =$ $\{ (x,y) \mid x=y+9 \}$

$[(1,10)] =$ $\{ (x,y) \mid x+9=y \}$

Whenever we have a congruence in an interpretation, we can shrink the structure to a smaller one by merging elements that are in congruence.

In the resulting structure $([N^2], I_Q)$ we define operations $p$ and $m$ such that the following holds:

$\begin{equation*} \begin{array}{l} I_Q(p)( [(x_1,y_1)] , [(x_2,y_2)] ) = [(x_1 + x_2, y_1 + y_2)] \\ I_Q(m)( [(x_1,y_1)] , [(x_2,y_2)] ) = [(x_1 + y_2, y_1 + x_2)] \end{array} \end{equation*}$

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 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

(Recall notation in First-Order Logic Semantics.)

Let $I = (D,\alpha)$ be an interpretation of language ${\cal L}$ with $eq \in {\cal L}$ for which Axioms for Equality hold, that is, $\alpha(eq)$ is a congruence relation for $I$ . We will construct a new model $I_Q$ .

For each element $x \in D$ , define

$\begin{equation*} [x] = \{ y \mid (x,y) \in \alpha(eq) \} \end{equation*}$

Let

$\begin{equation*} [D] = \{ [x] \mid x \in D \} \end{equation*}$

The constructed model will be $I_Q = ([D],\alpha_Q)$ where

$\begin{equation*} \alpha_Q(R) = \{ ([x_1],\ldots,[x_n]) \mid (x_1,\ldots,x_n) \in \alpha(R) \} \end{equation*}$

In particular, when $R$ is $eq$ we have

$\alpha_Q(eq) =$ $\{ ([x_1],[x_2]) \mid (x_1,x_2) \in \alpha(eq) \} = \{ (a,a) \mid a \in D \}$ that is, the interpretation of eq in $([D],I_Q)$ is diagonal relation - equality.

Functions are special case of relations:

$\begin{equation*} \alpha_Q(f) = \{ ([x_1],\ldots,[x_n],[x_{n+1}]) \mid (x_1,\ldots,x_n,x_{n+1}) \in \alpha(f) \} \end{equation*}$

Interpretation of variables is analogous to interpretation of constants:

$\begin{equation*} \alpha_Q(x) = [\alpha(x)] \end{equation*}$

Lemma 0: For all $x_1,\ldots,x_n \in D$ ,

$\begin{equation*} ([x_1],\ldots,[x_n]) \in \alpha_Q(R) \mbox{ iff } (x_1,\ldots,x_n) \in \alpha(R) \end{equation*}$

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 $x_1,\ldots,x_n \in D$ ,

$\begin{equation*} \alpha_Q(f)([x_1],\ldots,[x_n]) = [\alpha(f)(x_1,\ldots,x_n)] \end{equation*}$

Lemma 2: For each term $t$ we have $e_T(t)(I_Q) = [e_T(t)(I)]$ .

Theorem: For each formula $F$ that contains no '=' symbol, we have $e_F(F)(I) = e_F(F)(I_Q) = e_F(F_0)(I_Q)$ where $F_0$ is result of replacing 'eq' with '=' in $F$ .