In Tao's book Analysis 1, he writes:
Thus, from the point of view of logic, we can define equality on a [remark by myself: I think he forgot the word "type of object" here] however we please, so long as it obeys the reflexive, symmetry, and transitive axioms, and it is consistent with all other operations on the class of objects under discussion in the sense that the substitution axiom was true for all of those operations.
Does he mean that, if one wants to define define equality on a specific type of object (like functions, ordered pairs, for example), one has to check that these axioms of equality (he refers to these four axioms of equality as "symmetry", "reflexivity", "transitivity", and "substitution") hold in the sense that one has to prove them? It seems so, because of these two passages:
[In section 3.3 Functions] We observe that functions obey the axiom of substitution: if $x=x'$, then $f(x) = f(x')$ (why?).
(My answer would be "because that's an axiom", but Tao apparently wouldn't accept that.)
And after defining equality of sets ($A=B:\iff \forall x(x\in A\iff x\in B)$), Tao writes (on page 39):
One can easily verify that this notion of equality is reflexive, symmetric, and transitive (Exercise 3.1.1). Observe that if $x\in A$ and $A = B$, then $x\in B$, by Definition 3.1.4. Thus the "is an element of" relation $\in $ obeys the axiom of substitution
So he gives the exercise to prove the axioms of equality for sets. Why does one has to prove axioms? Or, put differently: if one can prove these things, why does he state them as axioms?
Best Answer
I believe Tao means that the axioms of reflexivity, symmetry, and transitivity are adequate to capture our pre-existing (this is the key) intuition about what "equality" between two objects ought to be. Let me try two contrasting examples to help unpack what I mean.
Version 1
You: Sets $A$ and $B$ are shmequal provided $x \in A \Leftrightarrow x \in B$ for all $x$.
Me: That sounds like a fine relation to investigate. Creative name, by the way.
Version 2
You: Sets $A$ and $B$ are equal provided $x \in A \Leftrightarrow x \in B$ for all $x$.
Me: Now, hold on just a second. By "equal", you mean "identical" or "exactly the same"? I'm not sure I'm ready to accept that this abstract definition captures all that. You would need to show me that the relation $x \in A \Leftrightarrow x \in B$ for all $x$ is reflexive, symmetric, and transitive before I'm willing to concede that this deserves a name like "equal".
Comment from OP
An example came to mind: we define equality for ordered pairs: $(x,y)=(a,b)⟺x=a∧y=b$. To show that equality for ordered pairs is reflexive we need to show $(x,y)=(x,y)$, which by definition means $x=x∧y=y$. But $x$ and $y$ could itself be ordered pairs. So now we are in the situation where we have to prove that every ordered pair is equal to itself but where we also have to accept this as given.
A weak response from me
I struggle to find good words to address your question, but it might help to remember that we are not checking whether $(x,y) = (x,y)$. Rather this is one of the things we insist should be the case if "equality" is to mean anything; it must apply to identical objects. Instead, we are asking "For ordered pairs, does the $x = a \wedge y = b$ property capture this self-evident truth about equality?". We find that it does: $x = x$ and $y = y$ are both true statements because we are comparing two identical objects in each case.