How to prove that the equality relation is an equivalence relation? Fraleigh's algebra book presents the following

Definition. Let $X$ be a  set. The equality relation in $X$ is the subset $$\{(x,x);\;x\in X\}\subset X\times X.$$

and also
0.19 Example of Fraleigh's algebra book says
For any nonemtpy set $X$, the equality relation $=$ defined by the subset $$\{(x,x);\;x\in X\}\subset X\times X.$$ is an equivalence relation.
I tried to prove that the equality relation is an equivalence relation. and I failed.
but Tao's analysis1 says in the appendix that equality just obeys the following four $axioms$
reflexive, symmetry, transitive, and substitution axioms.
so what is the truth?
Is the equality relation just an equivalence relation by axioms?
or
Is the equality relation provable that it is an equivalence relation?
 A: Reflexivity: For $x\in X$, we have $(x,x)$ in the equality relation by definition.
Symmetry: Suppose $(x,y)$ is in the equality relation. Then, by definition, $x=y$, so $y=x$; thus $(y,x)=(x,x)$ is in the equality relation.
Transitivity: Suppose $(x,y), (y,z)$ are in the equality relation. Then $x=y$ and $y=z$, so $x=z$. Hence $(x,z)=(x,x)$ is in the equality relation.
Thus the equality relation is an equivalence relation.
A: Is the equality relation just an equivalence relation by axioms?
or
Is the equality relation provable that it is an equivalence relation?
I think the answer depends on the precise meaning of "equality".
Definition 1. We say that "$x$ equals $y$" and write "$x=y$" provided that "$x$ is $y$", that is, "$x$ is identical with $y$", that is, "$x$ and $y$ designate the same entity". In this case, the expression $x=y$ is called an "equality" or an "identity".
In this context, reflexivity, symmetry and transitivity are axioms. In view of the given definition, they are quite natural because it is clear (from our intuition) that:

*

*everything is identical with itself.

*if $x$ is identical with $y$, then $y$ is identical with $x$.

*if $x$ is identical with a thing that is identical with $z$, then $x$ is identical with $z$.

This idea of equality (identity) arises when we are defining our rules of inference, which are the rules that allow us to prove our theorems. Specifically, at this level of the theory, we have to introduce the

Rule Governing Identities: If $S$ is an open formula, from $S$ and $t_1=t_2$, or from $S$ and $t_2=t_1$ we may derive $T$, provided that $T$ results from $S$ by replacing one or more occurences of $t_1$ in $S$ by $t_2$. Moreover, the identity $t=t$ is derivable from the empty set of premises. (Suppes book, Chapter 5)

This rule (at least according to the Suppes approach, which I believe is the most elementary and usual) is considered as part of our basic first-order predicate logic, where the symbol $=$ is treated as a universal logical symbol.
On the other hand, we have the
Definition 2. Given a set $X$ and $x,y\in X$, we say that "$x$ equals $y$" and write "$x=y$" provided that $xRy$, where $R$ is the relation defined by $R=\{(x,x)\mid x\in X\}$ and called "equality relation".
Here, the fact that the equality (equality relation) is reflexive, symmetric and transitive is a theorem. However, in order to prove it we need the rule governing identity (in particular, we need reflexivity, symmetry and transitivity of identity). It is allowed to use it because it is part of our basic inference rules. And the reasoning is not circular because we are using different ideas of "equality".
A: You can prove that the equality relation is an equivalence relation by proving that it fulfill reflexivity, symmetry and transitivity axioms.
A: Proof. Set $E=\{(x,x): x\in X\}$. We show it satisfies the three conditions one bye one.

*

*Reflexivity. For any $x\in X$, by the definition of the equality relation we have $xEx$.

*Symmetry. For any $x,y\in X$, if $xEy$, also by the definition of the equality relation we have $x=y$, and so we have $yEx$.

*Transitivity. For any $x,y,z\in X$, if $xEy$ and $yEz$, also by the definition of the equality relation we have $x=y=z$, and so we have $xEz$.


Therefore, it can be seen a provable theorem with no axioms as premises that the equality relation is an equivalence relation!
