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To put it briefly: In model theory, we are allowed to interpret any relation symbol in any way we like. So why do people seem to require that "$=$" is interpreted as the actual equality?

Let me elaborate a little more. In model theory, as I imperfectly understand it, one starts with an alphabet $\Sigma$ consisting of the allowable function and relation symbols; for instance for ordered fields we could take $\Sigma = \{\cdot,+,<,0,1\}$. To these, we add symbols for variables $x_1,x_2,\dots$ and logical symbols $\vee, \wedge, \forall, \exists, \dots$. Using these, we can form terms (well formed expressions that will describe elements of the set) and sentences (expressions that will either be true or false). If we assume some set $A$ of sentences to be true (axioms), then the set of all their logical consequences, say $T$, is a theory. A theory can be interpreted by first choosing a set $X$ to work on, and then assigning to the function and relation symbols actual functions ($X^k \to X$) and relations ($X^k \to \{\top,\bot\}$). This is to be done in such a way that the axioms are satisfied.

My problem is that the equality seems to be treated in a different way than other relations, and I don't quite see why. As far as I understand, it is normally required to be the "real" identity: $x = y$ means that $x$ and $y$ are the same element of $X$. Is there some reason not to treat "$=$" just as an ordinary relation (with axioms of being an equivalence relation + for each relation symbol axiom "if $x_1 = y_1,\dots,x_k=y_k$, then $R(x_1,\dots,x_k)$ iff $R(y_1,\dots,y_k)$" )? What would go wrong if we did?

(The reason I am asking is mostly that it seems to me that this would make some constructions more elegant (such as the ultraproducts) by eliminating a quotient.)

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    $\begingroup$ I think it's okay if we treat equality as just another relation. Some logic textbooks in fact define equality exactly as you did. $\endgroup$
    – Adriano
    Jun 1, 2013 at 23:29
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    $\begingroup$ You might find this wikipedia section satisfactory: en.wikipedia.org/wiki/… $\endgroup$
    – Denis
    Jun 1, 2013 at 23:29
  • $\begingroup$ @Adriano: Thank you! May I ask you for a reference to some of these textbooks? $\endgroup$
    – Jakub Konieczny
    Jun 1, 2013 at 23:31
  • $\begingroup$ In probability theory, $=$ is often not treated as equality. In particular, if $X$ and $Y$ are random variables on a common outcome space $\Omega$, then $X=Y$ is usually taken as shorthand for $\{\omega \in \Omega : X(\omega) = Y(\omega)\}.$ $\endgroup$
    – goblin GONE
    Dec 2, 2013 at 2:01

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It is not wrong. We can interpret '$=$' as an arbitrary binary relation symbol, then if we insert the statements to $A$ that it is equivalence relation and that it preserves all other relation and function symbols, then it will be interpreted as a congruence relation $\sim$ on a model $X$, and we can calmly form the quotient model $X/\sim$, which behaves exactly the same way as $X$, and '$=$' will be interpreted as real equality therein.

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To add a historical twist to the fine answers given earlier, note that Fermat, Leibniz, and Euler precisely considered a generalized notion of equality in their work. Fermat is discussed here: http://www.mitpressjournals.org/doi/abs/10.1162/POSC_a_00101, Leibniz here: http://www.ams.org/notices/201211/rtx121101550p.pdf, and Euler here: http://u.cs.biu.ac.il/~katzmik/bairetal.html

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When I learned logic in college, the equality relation was treated specially: rather than taking = as a possibly optional symbol in a particular language, it was a built-in feature that had the same syntactic role as any other relation symbol, but was required to have the interpretation of actual equality in a model. That is, it had the same status as $\neg$, which one could interpret as a unary relation (i.e. a predicate) but is in fact fixed in meaning by the rules of logic.

There is no disadvantage to having a symbol with a certain requisite meaning in every model, so long as that meaning is available regardless of the language. For one could otherwise just simulate it with a relation symbol, as you say, except that we'd get other (presumably spurious) models with = not meaning equality. Of course, since almost any set with any structures can be a model for some theory, there isn't much room for this kind of thing; however, equality is one relation you can guarantee exists, so we do.

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No, there is nothing wrong with treating "$=$" as just another relation, although that approach seems to be out of fashion. See Abraham Robinson's 1965 book Introduction to Model Theory and to the Metamathematics of Algebra. If memory serves (I don't have a copy) it treats equality the way you are asking for.

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I think that Leibnitz' identity is a part of the answer.

As I see it $'\!\!='$ is not just any equivalence relation, but the ultimate. Given any equivalence relation $'\!\!\sim '$, then it must be true that $x=y \Rightarrow x\sim y$.

Not all phenomena are identical to others or even to themselves: it is virtually impossible to define an identity for statements (here equivalence is the best, but obvious different statements can be equivalent) or even for some physical phenomena as photons and waves.

I think this is a question about how (for example) mathematical objects differs from other mathematical features. The question of the existence of mathematical objects, should rely on explicitly stated postulates or paradigms.

Also, the question of the identity is closely related to the principle of substitution in mathematics.

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