# Expressing interrelationships in first order logic

I'm trying to figure out how to best formalize the following interrelationship in first order logic:

A material has (electric) resistance $r$ and conductance $g$, and the two are related as $r \cdot g = 1$.

The idea is that, if I know either $r$ or $g$ for some material, I should be able to infer the other.

Let the relation $\mathrm{res}(m,r)$ mean that material $m$ has resistance $r$, and similarly $\mathrm{con}(m,g)$ for conductance, and let $\mathrm{mul}(x,y,z)$ denote the multiplication relation $x \cdot y = z$. The following formulas express the semantics of resistance/conductance:

1. $\mathrm{res}(m,r) \land \mathrm{con}(m,g) \rightarrow \mathrm{mul}(r,g,1)$
2. $\mathrm{res}(m,r) \land \mathrm{mul}(r,g,1) \rightarrow \mathrm{con}(m,g)$
3. $\mathrm{con}(m,g) \land \mathrm{mul}(r,g,1) \rightarrow \mathrm{res}(m,r)$

I think this works as expected: for example, if I know $\mathrm{res}(X,10)$ for some specific material $X$ (a constant), then rule (2) together with the fact $\mathrm{mul}(10,0.1,1)$ gives $\mathrm{con}(X,0.1)$. However, it feels like we should be able to express the relation between $r$ and $g$ more compactly$-$do we really need three formulas for this?

Intuitively, I thought there should be a single formula describing the relation, but I can't find one that works. Am I missing something, or is there something lurking here that's beyond the expressivity of first order logic?

## 2 Answers

Are you somehow not allowing your logical language to contain function symbols?

I would write $$\forall x(\operatorname{material}(x) \to \operatorname{res}(x)\cdot\operatorname{con}(x) = 1)$$ and be done with it.

(Even though first-order logic can be formulated with no function symbols without losing expressivity, the standard presentations allow any combination of predicate symbols, function symbols and constant symbols).

If I understood you correctly, this relationship could be written just as \begin{equation} res(m,r) \leftrightarrow con(m,\frac{1}{r}) \end{equation}

EDIT (after discussion in the comments): \begin{equation} (res(m,r) \land con(m,g)) \leftrightarrow mul(r,g,1) \end{equation} For a specific material (say $m=X$), we have the following;

• If $res(X,r)$ holds, $con(X,g)$ must hold for $g$ such that $mul(r,g,1)$ holds.
• If $con(X,g)$ holds, $res(X,r)$ must hold for $r$ such that $mul(r,g,1)$ holds.

I am aware that this lacks the restriction on $m$, but I figured that this formula is not to be used without knowing either $r$ or $g$ of a specific material $m$.

• Can you give some explanation? – user144410 Aug 19 '18 at 10:16
• Following what @Roland said: "..., if I know either r or g for some material, I should be able to infer the other". So, if we know the resistance of $m$ is $r$, than conductance of $m$ must be $1/r$ and vice-versa. Therefore I put logical equivalence, that covers both ways of implication. – Sandro Lovnički Aug 19 '18 at 10:20
• Well, that seems simple, but it doesn't look like first order logic. What exactly is the $1/r$ construct? If it's a function (taking terms to terms) then you cannot prove $res(X, 10)$ from $con(X, 0.1)$, because there's no way to unify $0.1$ with the function to get to the value of $r$. I'm pretty sure the inverse must be handled implicitly via a relation. – Roland Aug 19 '18 at 10:37
• You are right @Roland, it cannot be done with less than 3 relations. Can it be done if we add your $mul$ relation to above biconditional formula? What are the problems then? – Sandro Lovnički Aug 19 '18 at 11:01
• @SandroLovnički: Your proposal would seem to say that whenever we can find an $r$ and $g$ that multiply to $1$, those particular $r$ and $g$ will magically be parameters of every $m$ in the universe ... – hmakholm left over Monica Aug 19 '18 at 13:19