In first-order logic , the **theory of partial order** has the language $\mathscr{L}={\leq}$ and the following three axioms:

1. $(\forall xyz)((x\leq y\land y\leq z)\rightarrow x\leq z)$

2. $(\forall xy)((x\leq y \land y\leq x)\rightarrow x\equiv y)$

3. $(\forall x)(x\leq x)$

Every model of this theory consists of $\langle A,\leq\rangle$, where $A$ is a set partially ordered by $\leq$. The **theory of total order (linear order, simple order)** has the following additional axiom:

4. $(\forall xy)(x\leq y\lor y\leq x)$

which states, in plain English, that any two elements of the order are comparable - one must be greater than or equal to the other. Additional axioms may be added to place further restrictions on the types of orders under consideration. For instance, the **theory of dense total orders** has the following additional axioms:

5. $(\forall xy)(x < y \rightarrow (\exists z)(x<z\land z<y))$

6. $(\exists xy)(x\not\equiv y)$

where $x<y$ is shorthand for $x\leq y \land x\not\equiv y$.

*There is no theory of well-orders *. One of the exercises from Chang and Kiesler listed here is to show that the models $\langle \omega,\leq\rangle$ and $\langle \omega+\omega^d+\omega,\leq\rangle$ are elementarily equivalent. Thus, if the former is a model of any theory, so is the latter, and vice versa. But the former is a well-order and the latter is not, meaning that there cannot exist a theory whose models are precisely the well-ordered total orders. In fact, we can go even further with the following exercise from Chang and Kiesler:

Exercise 1. (Chang and Kiesler)Let $\mathfrak{A}=\langle A,\leq,...\rangle$ be an infinite model such that $\leq$ well-orders $A$. Show that there is a model $\mathfrak{A}'=\langle A',\leq',...\rangle$ such that $\leq'$ doesnotwell-order $A'$.

Solution.We will use what Chang and Kiesler call the "method of diagrams". Let $\mathfrak{A}^\ast$ be the expansion on $\mathfrak{A}$ to a larger language in which one constant element is added for each element of $A$ that is not already the interpretation of a constant, so that each element is the interpretation of exactly one constant. Further, let $c_0,c_1,c_2,...$ be a new set of constant symbols not already used in the new language, and let $\Delta$ be the set of sentences Let $\Sigma$ be the set of sentences that are true in $\mathfrak{A}'$. Then every finite subset of $\Sigma\cup\Delta$ is consistent, for each finite subsets only contains at most finitely many sentences in the form $c_{i+1}<c_i$, and if there are exactly $n$ such sentences, the constants used in these sentences can be consistently interpreted as elements of some finite suborder of $A$ of length $2n$. This means, by the compactness theorem, that $\Sigma\cup\Delta$ has a model. Let $\mathfrak{A}'$ be the reduct of this model to the original language without added constant symbols. This model is elementarily equivalent to $\mathfrak{A}$, but it is not well-ordered, because its elements corresponding to the interpretations of the constant symbols $c_i$ in the expanded model form an infinite descending chain. $\blacksquare$