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# 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 $a, b \in S$. $$a, b \in S \land aRb \equiv (\forall x \in S :: aRx \equiv bRx)$$