# Predicate Logic, formalization of an equivalence relation.

I was wandering how to define an equvalence relation using predicate logic, and suddently I realized that I was probably messing with somenthing relating open and closed formulas...

$\bf{Exercise \;}$: formalize in the language $\mathcal{L}=\{R\}\,$ where $R$ is the only symbol of relation "$R$ is an equivalence relation with at least two classes"

I'm confused about these two formulas:

$${\bf{(1)}}\;\; R(x,y)= R(x,x)\wedge (R(x,y)\longleftrightarrow R(y,x)) \,\wedge \,\forall z\,((R(x,y)\,\wedge\,R(y,z))\longrightarrow\,R(x,z))\;\wedge\;\;\;\;\;\;\;\;\;\; \exists \,v \,\exists \,w\,(R(w,v)\,\wedge \neg (R(w,x)\vee R(v,x))$$

It's an open formula, what does it say? it's Probably identifying a set of $(x,y)$ that has those properties.

$${\bf{(2)}}\;\; R= \forall x\,\forall y\,R(x,x)\wedge (R(x,y)\longleftrightarrow R(y,x)) \,\wedge \,\forall z\,((R(x,y)\,\wedge\,R(y,z))\longrightarrow\,R(x,z))\;\wedge\;\;\;\;\;\;\;\;\;\; \exists \,v \,\exists \,w\,(R(w,v)\,\wedge \neg (R(w,x)\vee R(v,x))$$

What's the difference?

Why the formula has to be closed?

1) is indeed an open formula with two free variables, $x$ and $y$. This, however, can only be used to express what it takes for $x$ and $y$ to stand in some relationship $R$ ... it cannot be used to express that R is an equivalence relationship, because that requires you to make a general claim about all pairs of objects in the domain.

So yes, 2) is the way to go. I would recommend a few minor changes though:

• remove the $R=..$ at the beginning, since that is not part of the logic expression

• either recycle the quantifiers for each of the reflexivity, symmetry, and transivity part, or have just one set of quantifiers ... you are somewhere in between. Personally, I would recycle the quantifiers so the whole expression becomes one big conjunction ... that way, you can immediately do a conjunction simplification to get the separate parts if needed for a formal proof

• you can greatly simplify the part about there being at least two classes. All you need is: $\exists x \exists y \, \neg R(x,y)$

So, in sum:

$$\forall x R(x,x) \; \land$$ $$\forall x \forall y (R(x,y)\leftrightarrow R(y,x)) \; \land$$ $$\forall x \forall y \forall z\,((R(x,y)\,\land\,R(y,z))\rightarrow\,R(x,z))\;\land$$ $$\exists x \exists y \; \neg R(x,y)$$