Question: Probably, it's a elementary question, but I would like some explanation.

Everyone knows that equality relation is (i) reflexive, (ii) symmetric and (iii) transitive, that is, satisfies

(i) $x=x$;  

(ii) $x=y\Rightarrow y=x$;  

(iii) $x=y$ and $y=z\Rightarrow x=z$.

I'm interested in to deduce these properties from some appropriate definition of equality. Fraleigh's algebra book presents the following

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

Let $\mathcal{R}$ be the equality relation in $X$.

By definition, $x\mathcal{R}x$ for all $x\in X$. So, $\mathcal{R}$ is reflexive.

**Question 1.** How to write a formal proof to the symmetry and the reflexivity?

Fraleigh denotes the equality relation by $=$ and, after the definition, says "Thus for any $x\in X$, we have $x=x$ but if $x$ and $y$ are different elements of $X$, then $(x,y)\notin =$ and we write $x\neq y$".

**Question 2.** Has the above comment any sense? For me, it sounds like "if $(x,y)\notin =$ then $(x,y)\notin =$".

Thanks.

Answer: 
This looks trivial, but if it needs a proof it isn't - you can't rely on trivialities then. It depends on actual definitions and the proof will look different depending on the exact definitions.

It all depends on the definition of your set-building notation. The notation $R=\{(x,x): x\in X\}$ actually depends on the notion of equality (of pairs) already. It means that $(u,v) \in R\Leftrightarrow (u,v)\in X^2 \land \exists x\in X: (u,v)=(x,x)$ (or do you have another definition?). It then falls back on the corresponding property of the equality notation (not the same as the equality relation $R$).

For example $(u,v) \in R$ means that there's an $x$ such that $u=x$ and $v=x$, but to get from there to $u=v$ you need to use symmetry and transitivity of $=$ (ie that $v=x\Rightarrow x=v$ and $u=x\land x=v\Rightarrow u=v$).

The reflexivity, symmetry and transitivity of equality follows from it's definition $u=v \Leftrightarrow (\forall \xi:\xi \in u \Leftrightarrow \xi\in v)$ and it will be reduced to the similar property of the $\Leftrightarrow$ notation. That is $p \Leftrightarrow p$, $(p \Leftrightarrow q) \Rightarrow (q \Leftrightarrow p)$ and $(p\Leftrightarrow q) \land (q\Leftrightarrow r) \Rightarrow (p\Leftrightarrow r)$
So the equality relation just means that things are "equal" when they're the same (like $x$ and $x$) and "not equal" when they're not the same (like $x$ and $y$).

You've already shown it's reflexive.

For symmetry, suppose that $$x=y$$ Then $$x=x$$ so "switching" :) :) $$y=x$$

For transitivity, suppose $x=y$ and $y=z$. Then $$x=y=z$$ so $$x=z$$ This is one of those proofs that seems so obvious its more confusing to just not formally prove it.

EDIT: on second thought, the best way to think about it is as a partition. The "equality relation" partitions each element into their own private cell. How anti-social.