Is this alternative definition of 'equivalence relation' correct?

I was puzzling over another question: Let $R$ be an equivalence relation on a set $A$, $a,b \in A$. Prove $[a] = [b]$ iff $aRb$. And this made me discover that $$(0) \; \langle \forall a,b :: aRb \equiv \langle \forall x :: aRx \equiv bRx \rangle \rangle$$
is an alternative characterization of $R$ being an equivalence relation, and one which only uses logical equivalence. (Note that throughout this question, $a$, $b$, $c$, and $x$ implicitly range over a fixed set $A$.)

Update 3: To put it more explicitly and longwindedly, I discovered that $$R \textrm{ is an equivalence relation on } A \;\equiv\; \langle \forall a,b \in A :: aRb \:\equiv\: \langle \forall x \in A :: aRx \equiv bRx\rangle \rangle$$

(I’ve updated this question to ask for the correctness of this definition, and have posted my original question as a separate one: Is this alternative definition of 'equivalence relation' well-known? useful? used?)

The nice thing about this definition is that it says, “equivalence of $a$ and $b$ means that it doesn’t matter which of the two you substitute”, which seems to be the essence of equivalence (cf. Leibniz’ law).

Is this alternative definition well-known? useful? used? correct?


Update: Two comments and one answer suggest that my $(0)$ is not equivalent to the combination of
\begin{align}
(1) & \langle \forall a :: aRa \rangle \\
(2) & \langle \forall a,b :: aRb \;\Rightarrow\; bRa \rangle \\
(3) & \langle \forall a,b,c :: aRb \land bRc \;\Rightarrow\; aRc \rangle \\
\end{align}
However, as far as I can see these are equivalent. One direction is proven in the answers to the question referenced above. For the other direction, we have for all $a$
\begin{align}
& aRa \\
\equiv & \;\;\;\;\;\text{“using $(0)$ with $b:=a$”} \\
& \langle \forall x :: aRx \equiv aRx \rangle \\
\equiv & \;\;\;\;\;\text{“logic”} \\
& \textrm{true} \\
\end{align}
and for all $a$ and $b$
\begin{align}
& aRb \\
\equiv & \;\;\;\;\;\text{“by $(0)$”} \\
& \langle \forall x :: aRx \equiv bRx \rangle \\
\Rightarrow & \;\;\;\;\;\text{“choose $x:=a$”} \\
& aRa \equiv bRa \\
\equiv & \;\;\;\;\;\text{“by $(1)$”} \\
& bRa \\
\end{align}
and for all $a$, $b$, and $c$
\begin{align}
& aRb \land bRc \\
\equiv & \;\;\;\;\;\text{“by $(0)$, twice”} \\
& \langle \forall x :: aRx \equiv bRx \rangle \land \langle \forall x :: bRx \equiv cRx \rangle \\
\Rightarrow & \;\;\;\;\;\text{“logic”} \\
& \langle \forall x :: aRx \equiv cRx \rangle \\
\equiv & \;\;\;\;\;\text{“by $(0)$ with $b:=c$”} \\
& aRc \\
\end{align}

Or am I missing something?


Update 2: In the above I have now made all quantifications explicit, to prevent any confusion.

Solutions Collecting From Web of "Is this alternative definition of 'equivalence relation' correct?"

What you have almost works. The only problem with it is that it makes $R$ a proper class, since it implies that $xRx$ for all sets $x$. You really ought to quantify over the domain of $R$; if that’s $D$, you want

$$\forall x,y\Big(xRy\leftrightarrow x,y\in D\land\forall z\in D(xRz\leftrightarrow yRz)\Big)\;.$$

Now you get $xRx$ precisely for those sets $x$ that are elements of $D$, which is what you want.

Your characterization of $R$ may “beg the question”: the characterization of an equivalence relation:$$aRb \equiv \langle \forall x :: aRx \equiv bRx \rangle$$

works fine, provided we know that $R$ is reflexive. So if you’re trying to define an equivalence relation, we run into problems.

However, if you have the added premise: “given $R$ is an equivalence relation,” then indeed, $R$ is reflexive.
So if you are offering an alternative characterization of what it means for two elements to be related under $R$, given that $\,R\,$ is an equivalence relation, then I think your characterization is fine, and is appealing at an intuitive level. However, you’ll want to restrict $x$ to the domain (set $S$) on which the equivalence relations $R$ is defined, and also add the qualification that $a, b \in S$. $$a, b \in S \land aRb \equiv (\forall x \in S :: aRx \equiv bRx)$$