# What is the intuition behind the negative exponential '?' in linear logic?

The positive exponential $$!$$ has a very satisfying interpretation in terms of the standard resource interpretation of linear logic. Given a resource $$a$$, we know that $$!a$$ means an infinite supply of $$a$$. Or, stated more concretely in terms of the connectives of linear logic: $$!a \equiv !a \otimes a$$. My question is, under this same interpretation of atomic propositions $$a$$ in linear logic as resources, how are we to interpret $$?a$$?

The best interpretation I've seen so far seems to be that $$?a$$ "consumes $$a$$'s", but what does that mean concretely in terms of the logic? Is there an analogue of the formula $$!a = !a \otimes a$$ for $$?$$?

Unfortunately, the usual sequent rules for linear logic are not very helpful to my intuition. I think the two-sided sequent presentation of linear logic is the most intuitive so far — $$!$$ reintroduces weakening/contraction on the left side of a sequent, and $$?$$ allows for the same on the right side of a sequent (as evidenced by the rules $$!W, ?W, !C, ?C, !D,$$ and $$?D$$ in the presentation here). But what are we to make of the rules $$?L$$ and $$!R$$?

So far, I have not been able to find an adequate or complete explanation of either the resource interpretation of $$?$$ in general, or even the meaning of the rules $$?L$$ and $$!R$$, whether it be with regards to the resource interpretation, or the original motivations of linear logic by Girard. And I am curious whether anyone could point me towards better explanations or resources than I have been able to find.

The key to understanding the negative exponential is first understanding the interpretation of negation in linear logic. Unlike in classical logic, where negation roughly indicates the absence of something (i.e. the absence of truth), this does not carry over to the resource interpretation of linear logic. $A^{\perp}$ is best thought of as not the absence of an $A$, but the demand for an $A$. And hence, when we have in our possession both an $A$ and an $A^\perp$, we obtain $1$, or the absence of resources. In other words, $A \otimes A^\perp \multimap 1$.

Now that we have this, recall that $?A$ may be defined dually in terms of $!$ as $?A = !A^\perp$. Thus, in the resource interpretation of linear logic, $?A$ means infinite potential demand for the resource A.

• In the last paragraph of your answer, the proposed definition of ? needs another negation; it should read $?A=(!(A^\bot))^\bot$. Jun 29, 2016 at 22:07
• Ah, yes. Thanks for the correction. This complicates my interpretation then, I'll have to think about this more... Jun 29, 2016 at 22:56
• @Sintrastes Did you find a better interpretation ? May 20, 2017 at 18:02

The way I think of $$!A$$ is as a choice of how many $$A$$'s you want to receive, rather than simply an infinite number of them. You could write it as an additive disjunction like so:

$$!A \approx 1\&(A)\&(A\otimes A)\&(A\otimes A\otimes A)\& ...$$

The dual of that would be:

$$\DeclareMathOperator{\par}{\unicode{8523}} ?A = (!A^\bot)^\bot \approx \bot\oplus(A)\oplus(A\par A)\oplus(A\par A\par A)\oplus ...$$

So $$?A$$ is sort of similar to an unknown number (zero or more) of $$A$$'s that must be used in parallel in that way that $$\par$$ requires.

Edit: This interpretation also explains a lot of the exponential sequents, and their rules start to look very similar to those for the additives. I'll just show the right-handed sequents, as described by the Stanford Encyclopedia of Philosophy:

Weakening: If you can get $$\Gamma$$, you can also produce some number (i.e. zero) of $$B$$. $$\cfrac{\Delta\vdash\Gamma}{\Delta\vdash?B,\Gamma}$$ Dereliction: If you can get $$B$$, you can instead produce some number (i.e. one) of $$B$$. $$\cfrac{\Delta\vdash B,\Gamma}{\Delta\vdash?B,\Gamma}$$ Contraction: If you have an unknown quantity of $$B$$ and another unknown quantity of $$B$$, you can join them into one unknown quantity. $$\cfrac{\Delta\vdash?B,?B,\Gamma}{\Delta\vdash?B,\Gamma}$$ Promotion: If you can use unlimited $$\Delta$$ to produce some number of $$\Gamma$$, and you can also produce a single $$B$$, then you can produce any amount of $$B$$ that you want. $$\cfrac{!\Delta\vdash B,?\Gamma}{!\Delta\vdash!B,?\Gamma}$$