TITLE: Example of first order logic without equality.
QUESTION [6 upvotes]: Most logic texts say that = is a special symbol which is always part of our language. It is my understanding, though, that it is perfectly acceptable to consider = to be an ordinary  binary relation or even to not include it.
My questions are:
Are there any example of times when it is beneficial to use a first order logic without the = symbol? (I am guessing no)
If the answer to the first question is no, then has anyone worked on these types of logics despite their apparent uselessness?

REPLY [5 votes]: Here are two examples where it is useful to omit equality. 
Second-order arithmetic
In the context of second-order arithmetic, we work with a two-sorted first order logic. There are two sorts of objects: "natural numbers" and "sets of natural numbers".  We include the equality sign for numbers, but not for sets of numbers. Instead, equality for sets of numbers is viewed as an abbreviation:
$$
X = Y \quad \text{means} \quad  (\forall n)(n \in X \leftrightarrow n \in Y).
$$
The motivation for this is that equality of natural numbers is decidable in an intuitive way: given two concrete numbers, we can tell whether they are the same. Also, telling whether a given number is or is not in a given set is viewed as decidable. But telling whether two (possibly infinite) sets of numbers are equal is much more difficult. There are particular results about the difficulty of deciding atomic formulas in models of second order arithmetic that would not hold if we included equality for sets in the basic language. 
Set theory
A second example is in traditional axiomatizations of set theory, which include only the membership symbol, $\in$, not the equality symbol. Equality is defined from $\in$: two sets are equal if they have the same members. In this setting, the motivation for removing equality is to be more parsimonious with basic concepts. In the traditional language of set theory, there is only one undefined symbol, $\in$, rather than two undefined symbols $\in$ and $=$.