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Consider the two structures $(\mathbb{R};f)$ and $(\mathbb{R};-)$, where $f$ represents additive inverse, and $-$ represents binary subtraction. Certainly, $f$ can be defined from $-$. Can $-$ be defined by $f$? I don't think it can, but I want a formal proof that no possible definition works.

Edit: It is very clear to me that $-$ can't be expressed by a term in $f$, but I am wondering if the graph of $-$ can be defined by a first-order formula in $f$.

user107952
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    Do you mean without having a $+$? – Randall Jul 13 '21 at 01:43
  • Can you clarify what you mean by "define"? – Eric Wofsey Jul 13 '21 at 01:53
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    In any case, for any reasonable notion of "define", the answer will be no, since $(\mathbb{R};f)$ has automorphisms that do not preserve $-$ (you should be able to find some for yourself). – Eric Wofsey Jul 13 '21 at 02:03
  • I am not sure that "$f$ can be defined from $-$" either. (Unless you specify "definable with parameters".) In fact, $-$ is translation invariant, $f$ is not. – Primo Petri Jul 13 '21 at 13:08
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    @PrimoPetri We have that $x - y = z$ implies $(x+5) - (y+5) = z$ (which is probably what you mean), but $(x+5) - (y+5) \neq z + 5$. So translation is not an automorphism. In fact, $f$ is $0$-definable in $(\mathbb{R}; -)$, because $0$ is $0$-definable in that structure (that's a lot of $0$s, but hopefully it's clear which means what). – Mark Kamsma Jul 13 '21 at 15:20
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    @PrimoPetri The term $(x-x)-x$ defines $f$. – user107952 Jul 13 '21 at 19:11

1 Answers1

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To move this off the unanswered queue, I'll expand Eric Wofsey's comment into an answer:

Definable structure is always preserved by automorphisms. To show that subtraction can't be defined from inversion alone, we just need to find some permutation of $\mathbb{R}$ respecting inversion but not respecting subtraction. One example which I quite like is the function $$b(x)=\begin{cases} x & \mbox{ if } x\in\mathbb{Q},\\ f(x) & \mbox{ if } x\not\in\mathbb{Q}.\\ \end{cases}$$ To see that $b$ does not in fact preserve subtraction just note that e.g. $b(\pi-(\pi-1))=b(1)=1$ but $b(\pi)-b(\pi-1)=-\pi-(1-\pi)-1$.

There are two quick notes worth making at this point:

  • First, the above analysis has very little to do with first-order logic. Isomorphism invariance of satisfaction, and consequently automorphism invariance of definability, is (almost always) one of the defining criteria for something to be a logic in the first place. So really what the above shows is that inversion can't define subtraction in any logical system.

  • Second, note that inversion is really combinatorial rather than algebraic when taken on its own: the structure $(\mathbb{R};f)$ is better thought of as just a set equipped with an equivalence relation, namely $\vert x\vert=\vert y\vert$, which partitions it into continuum-many pairs and one singleton. Thought of this way it's clear that we shouldn't expect any expressive power (and moreover this motivates non-interpretability, rather than mere non-definability, results).

Noah Schweber
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