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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?)

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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.

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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 $R$,

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