On decidability and axiomatizability of some ordered structures

Abstract

The ordered structures of natural, integer, rational and real numbers are studied here. It is known that the theories of these numbers in the language of order are decidable and finitely axiomatizable. Also, their theories in the language of order and addition are decidable and infinitely axiomatizable. For the language of order and multiplication, it is known that the theories of $\mathbb {N}$ and $\mathbb {Z}$ are not decidable (and so not axiomatizable by any computably enumerable set of sentences). By Tarski’s theorem, the multiplicative ordered structure of $\mathbb {R}$ is decidable also; here we prove this result directly and present an axiomatization. The structure of $\mathbb {Q}$ in the language of order and multiplication seems to be missing in the literature; here we show the decidability of its theory by the technique of quantifier elimination, and after presenting an infinite axiomatization for this structure, we prove that it is not finitely axiomatizable.

Appendix

Theorem 1

. The finite theory $\{\texttt {O}_1,\texttt {O}_2,\texttt {O}_3, \texttt {O}_4,\texttt {O}_5,\texttt {O}_6\}$ (of dense linear orders without endpoints—see Definitions 1 and 2) completely axiomatizes the theory of $\langle \mathbb {R};<\rangle$ and $\langle \mathbb {Q};<\rangle$, and so these structures are decidable. Moreover, (the theory of) those structures admit quantifier elimination.

Proof

All the atomic formulas are either of the form $u<v$ or $u=v$ for some variables u and v. If both of the variables are equal, then $u<u$ is equivalent with $\bot$ by $\texttt {O}_1$ and $u=u$ is equivalent with $\top$. So, by Remark 1, it suffices to eliminate the quantifier of the formulas of the form

(2)

where $y_i$’s, $z_j$’s and $u_k$’s are variables. Now, if $n\ne 0$ then formula (2) is equivalent with the quantifier-free formula

So, let us suppose that $n=0$. Then if $m=0$ or $\ell =0$ formula (2) is equivalent with the quantifier-free formula $\top$, by the axioms $\texttt {O}_5$ and $\texttt {O}_6$ (with $\texttt {O}_2$ and $\texttt {O}_3$) respectively, and if $\ell ,m\ne 0$ it is equivalent with the quantifier-free formula by the axiom $\texttt {O}_4$ (with $\texttt {O}_2$ and $\texttt {O}_3$). $\square$

Theorem 2

The finite theory of discrete linear orders without endpoints, consisting of the axioms $\texttt {O}_1$, $\texttt {O}_2$, $\texttt {O}_3$, $\texttt {O}_7$ plus

$$\begin{aligned} (\texttt {O}_8) \; \forall x \exists y (\mathfrak {s}(y)=x) \end{aligned}$$

completely axiomatizes the order theory of the integer numbers, and so its theory is decidable. Moreover, the structure $\langle \mathbb {Z};<,\mathfrak {s}\rangle$ admits quantifier elimination.

Proof

We note that all the terms in the language $\{<,\mathfrak {s}\}$ are of the form $\mathfrak {s}^n(y)$ for some variable y and $n\in \mathbb {N}$. So, all the atomic formulas are either of the form $\mathfrak {s}^n(u)=\mathfrak {s}^m(v)$ or $\mathfrak {s}^n(u)<\mathfrak {s}^m(v)$ for some variables u, v. If a variable x appears in the both sides of an atomic formula, then we have either $\mathfrak {s}^n(x)=\mathfrak {s}^m(x)$ or $\mathfrak {s}^n(x)<\mathfrak {s}^m(x)$. The formula $\mathfrak {s}^n(x)=\mathfrak {s}^m(x)$ is equivalent with $\top$ when $n=m$ and with $\bot$ otherwise; also $\mathfrak {s}^n(x)<\mathfrak {s}^m(x)$ is equivalent with $\top$ when $n<m$ and with $\bot$ otherwise. So, it suffices to consider the atomic formulas of the form $t<\mathfrak {s}^n(x)$ or $\mathfrak {s}^n(x)<t$ or $\mathfrak {s}^n(x)=t$ for some x-free term t and $n\in \mathbb {N}^+$. Now, by Remark 1, we eliminate the quantifier of the formulas

(3)

The axioms prove the equivalences $[a<b] \leftrightarrow [\mathfrak {s}(a)<\mathfrak {s}(b)]$ and $[a=b] \leftrightarrow [\mathfrak {s}(a)=\mathfrak {s}(b)]$; so we can assume that $p_i$’s and $q_j$’s and $r_k$’s in formula (3) are equal to each other, say to $\alpha$. Then by $\texttt {O}_8$ formula (3) is equivalent with

(4)

for some (possibly new) terms $t_i',s_j',u_k'$ (and $y=\mathfrak {s}^\alpha (x)$). Now, if $n\ne 0$ then formula (4) is equivalent with the quantifier-free formula

Let us then assume that $n=0$. The formula

(5)

is equivalent, by the axiom $\texttt {O}_7$ (in Definition 2), with the quantifier-free formula . $\square$

Theorem 4

The following infinite theory (of non-trivial ordered divisible abelian groups) completely axiomatizes the order and additive theory of the real and rational numbers, and so their theories are decidable. Moreover, the structures $\langle \mathbb {R};<,+,-,\mathbf{0}\rangle$ and $\langle \mathbb {Q};<,+,-,\mathbf{0}\rangle$ admit quantifier elimination.

$$\begin{aligned}&(\texttt {O}_1) \; \forall x,y(x<y\rightarrow y\not<x) \\&(\texttt {O}_2) \; \forall x,y,z (x<y<z\rightarrow x<z) \\&(\texttt {O}_3) \; \forall x,y (x<y \vee x=y \vee y<x) \\&(\texttt {A}_1) \; \forall x,y,z\,(x+(y+z)=(x+y)+z) \\&(\texttt {A}_2) \; \forall x (x+\mathbf {0}=x) \\&(\texttt {A}_3) \; \forall x (x+ (-x)=\mathbf {0}) \\&(\texttt {A}_4) \; \forall x,y (x+y=y+x) \\&(\texttt {A}_5) \; \forall x,y,z(x<y\rightarrow x+z<y+z) \\&(\texttt {A}_6) \; \exists y (y\ne \mathbf{0}) \\&(\texttt {A}_7) \; \forall x\exists y(x=n\centerdot y) \qquad \qquad \qquad n\in \mathbb {N}^+ \\ \end{aligned}$$

Proof

Firstly, let us note that $\texttt {O}_4$, $\texttt {O}_5$ and $\texttt {O}_6$ can be proved from the presented axioms: if $a<b$ then by $\texttt {A}_7$ there exists some c such that $c+c=a+b$; one can easily show that $a<c<b$ holds. Thus, $\texttt {O}_4$ is proved; for $\texttt {O}_5$ note that for any $\mathbf{0}<a$ we have $a<a+a$ by $\texttt {A}_5$. A dual argument can prove the axiom $\texttt {O}_6$. Also, the equivalences

• (i) $[a<b] \leftrightarrow [n\centerdot a< n\centerdot b]$ and

• (ii) $[a=b] \leftrightarrow [n\centerdot a=n\centerdot b]$

can be proved from the axioms: (i) follows from $\texttt {A}_5$ (with $\texttt {O}_1,\texttt {O}_2,\texttt {O}_3$) and (ii) follows from $\forall x (n\centerdot x=\mathbf{0}\rightarrow x=\mathbf{0})$ which is derived from $\texttt {A}_5$ (with $\texttt {O}_1,\texttt {O}_2,\texttt {O}_3$).

Secondly, every term containing x is equal to $n\centerdot x + t$ for some x-free term t and $n\!\in \!\mathbb {Z}\!-\!\{0\}$. So, every atomic formula containing x is equivalent with $n\centerdot x \Box t$ where $\Box \!\in \!\{=,<,>\}$. Whence, by Remark 1, it suffices to prove the equivalence of the formula

(6)

with a quantifier-free formula. By the equivalences (i) and (ii) above we can assume that $p_i$’s and $q_j$’s and $r_k$’s in formula (6) are equal to each other, say to $\alpha$. Then by $\texttt {A}_7$ formula (6) is equivalent with

(7)

for some (possibly new) terms $t_i',s_j',u_k'$ (and $y=\alpha \centerdot x$). Now, the quantifier of this formula can be eliminated just like the way that the quantifier of formula (2) was eliminated in the proof of Theorem 1. $\square$

About the congruence relations, a useful fact is the following generalized Chinese remainder theorem; below we present a proof of this theorem from (Fraenkel 1963).

Proposition 2

(Generalized Chinese remainder) For integers $n_0,n_1,\ldots ,n_k\geqslant 2$ and $t_0,t_1,\ldots ,t_k$ there exists some x such that $x\equiv _{n_i} t_i$ for $i=0,\ldots ,k$ if and only if $t_i \equiv _{d_{i,j}} t_j$ holds for each $0\leqslant i<j\leqslant k$, where $d_{i,j}$ is the greatest common divisor of $n_i$ and $n_j$.

Proof

The ‘only if’ part is easy. We prove the ‘if’ part by induction on k. For $k=0$ there is nothing to prove, and for $k=1$ we note that by Bézout’s Identity there are $a_0,a_1$ such that $a_0n_0 + a_1n_1 = d_{0,1}$. Also, by the assumption there exists some c such that $t_0-t_1=cd_{0,1}$. Now, if we take x to be $a_0(n_0/d_{0,1})t_1 + a_1(n_1/d_{0,1})t_0$ then we have $x=t_0-a_0n_0c$ and $x=t_1+a_1n_1c$ so $x\equiv _{n_0}t_0$ and $x\equiv _{n_1}t_1$ hold. For the induction step ($k+1$) suppose that $x\equiv _{n_i} t_i$ holds for $i=0,\ldots ,k$ (and that $t_i \equiv _{d_{i,j}} t_j$ holds for each $0\leqslant i<j\leqslant k+1$). Let n be the least common multiplier of $n_0,\ldots ,n_k$; then the greatest common divisor m of n and $n_{k+1}$ is the least common multiplier of $d_{0,k+1},\ldots ,d_{k,k+1}$. Now $x\equiv _{d_{i,k+1}}t_i$ holds for $0\leqslant i\leqslant k$ and so by the assumption $t_i\equiv _{d_{i,k+1}}t_{k+1}$ we have $x\equiv _{d_{i,k+1}}t_{k+1}$ (for $i=0,\ldots ,k$). Therefore, $x\equiv _m t_{k+1}$ and so $x-t_{k+1}=mc$ for some c. By Bézout’s Identity there are a, b such that $an+bn_{k+1}=m$. Now, for $y=x-anc$ we have $y=t_{k+1}+bn_{k+1}c\equiv _{n_{k+1}}t_{k+1}$ and also $y\equiv _{n_i}x\equiv _{n_i}t_i$ holds for each $0\leqslant i\leqslant k$. This proves the desired conclusion. $\square$

Theorem 5

The infinite theory of non-trivial discretely ordered abelian groups with the division algorithm, that is $\texttt {O}_1$, $\texttt {O}_2$, $\texttt {O}_3$, $\texttt {A}_1$, $\texttt {A}_2$, $\texttt {A}_3$, $\texttt {A}_4$, $\texttt {A}_5$ and

completely axiomatizes the order and additive theory of the integer numbers, and so its theory is decidable. Moreover, the (theory of the) structure $\langle \mathbb {Z};<,+,-,\mathbf{0},\mathbf{1},\{\equiv _n\}_{n\geqslant 2}\rangle$ admits quantifier elimination.

Proof

The axiom $\texttt {A}_7^\circ$ can be easily seen to be equivalent with the formula and so the negation signs behind the congruences can be eliminated by . Whence, by Remark 1, it suffices to show the equivalence of

(8)

with some quantifier-free formula, where $a_i$’s, $b_j$’s, $c_k$’s and $d_\ell$’s are natural numbers and $t_i$’s, $u_j$’s, $v_k$’s and $w_\ell$’s are x-free terms. By the equivalences

• (i) $[a<b] \leftrightarrow [n\centerdot a< n\centerdot b]$,

• (ii) $[a=b] \leftrightarrow [n\centerdot a=n\centerdot b]$ and

• (iii) $[a\equiv _m b] \leftrightarrow [n\centerdot a\equiv _{nm}n\centerdot b]$

which are provable from the axioms, we can assume that $a_i$’s, $b_j$’s, $c_k$’s and $d_\ell$’s in formula (8) are equal to each other, say to $\alpha$. Now, (8) is equivalent with

(9)

for $y=\alpha \centerdot x$ and some (possibly new) terms $t_i'$’s, $u_j'$’s, $v_k'$’s and $w_\ell '$’s. If $r\ne 0$ then (9) is readily equivalent with the quantifier-free formula which results from substituting $w_0'$ with y. So, it suffices to eliminate the quantifier of

(10)

By the equivalence of the formula $\exists x(\theta (x)\wedge u_0\!<\!x \wedge u_1\!<\!x)$ with the following formula

$$\begin{aligned} \big [\exists x (\theta (x)\wedge \!u_0\!<\!x)\!\wedge \!u_1\!\leqslant \!u_0\big ] \!\vee \!\big [\exists x (\theta (x)\!\wedge \!u_1\!<\!x)\!\wedge \!u_0\!\leqslant \!u_1\big ], \end{aligned}$$

we can assume that $p\leqslant 1$ (and $q\leqslant 1$ by a dual argument). Also, the formula $\exists x(\theta (x)\wedge x\equiv _{n_0}t_0 \wedge x\equiv _{n_1}t_1)$ is equivalent with $\exists x(\theta (x)\wedge x\equiv _{n}t) \wedge t_0\equiv _{d}t_1$ where d is the greatest common divisor of $n_0$ and $n_1$, n is their least common multiplier, and $t=a_0(n_0/d)t_1 + a_1(n_1/d)t_0$ where $a_0,a_1$ satisfy Bézout’s Identity $a_0n_0 + a_1n_1 = d$ (see the proof of Proposition 2). So, we can assume that $m\leqslant 1$ as well. Now, if $m=0$ then formula (10) is equivalent with a quantifier-free formula by Theorem 2 (with $\mathfrak {s}(x)=x+\mathbf{1}$ just like the way formula (5) was equivalent with some quantifier-free formula). So, suppose that $m=1$. In this case, if any of p or q is equal to 0 then (10) is equivalent with $\top$ (since any congruence can have infinitely large or infinitely small solutions). Finally, if $p=q=1=m$ then the formula

$$\begin{aligned} \exists x (x\equiv _{n}t \;\wedge \; u\!<\!x \;\wedge \; x\!<\!v) \end{aligned}$$

is equivalent with $\exists y (r<n\centerdot y\leqslant s)$ for $x=t+n\centerdot y$, $r=u-t$ and $s=v-t-\mathbf{1}$. Now, $\exists y (r<n\centerdot y\leqslant s)$ is equivalent with the quantifier-free formula

since by the division algorithm there are some q and some $i<n$ such that $s=qn + i$. The existence of some y such that $r<ny\leqslant s$ is then equivalent with $r<nq$ ($=s-i$). $\square$