Is currying valid in Linear logic and Relevance logic?

Question

In Classical and Intuitionistic logic we have what I will call the "currying equivalence":

$P \rightarrow (Q \rightarrow P) \equiv (P \land Q) \rightarrow P$

But linear and relevance logics do not allow weakening. To prove $P \rightarrow (Q \rightarrow P)$ we require weakening and the deduction theorem:

$$P \vdash P$$ $$P, Q \vdash P \hspace{0.3cm}$$ $$P \vdash Q \rightarrow P$$ $$\vdash P \rightarrow (Q \rightarrow P)$$

Is Currying invalid in Linear and Relevance logics?

SEP on Linear logic and Relevance logic.

Answer 1

In general, proving that two formulas $A$ and $B$ are provably equivalent in a logic system $S$ does not depend on the fact that $A$ and $B$ are provable in $S$. You might find two formulas that are not provable but are provably equivalent in a logic system. For example, every propositional variable $X$ is provably equivalent to $X$ in every coherent logic system $S$, even if $X$ is not provable in $S$. Clearly, you can't find two provably equivalent formulas $A$ and $B$ such that $A$ is provable and $B$ is not.

In linear logic (actually, already in its intuitionistic fragment) the "linear currying equivalence" holds, i.e. $P \multimap (Q \multimap P) \equiv (P \otimes Q) \multimap P$ (with respect to the exportation law in your original question, in linear logic you have to replace the implication with the linear implication $\multimap$, and the conjunction with the tensor $\otimes$). You can prove that using the rules of linear logic sequent calculus as follows: \begin{align*} \dfrac{\dfrac{\dfrac{\dfrac{}{P \vdash P} \qquad \dfrac{\dfrac{}{Q \vdash Q} \qquad \dfrac{}{P \vdash P}}{Q \multimap P, \, Q \vdash P}}{P \multimap (Q \multimap P), \, P, \, Q \vdash P}}{P \multimap (Q \multimap P), \, P \otimes Q \vdash P}}{P \multimap (Q \multimap P) \vdash (P \otimes Q) \multimap P} \qquad\qquad \dfrac{ \dfrac{ \dfrac{ \dfrac{\dfrac{}{P \vdash P} \qquad\dfrac{}{Q \vdash Q}}{P, \, Q \vdash P \otimes Q} \qquad \dfrac{}{P \vdash P} } {(P \otimes Q) \multimap P, \, P, \, Q \vdash P}} {(P \otimes Q) \multimap P, \, P \vdash Q \multimap P}} {(P \otimes Q) \multimap P \vdash P \multimap (Q \multimap P)} \end{align*}

This fact is not surprising: you can prove, for instance, $P \multimap (Q \multimap P)$ from $(P \otimes Q) \multimap P$ in a "linear" way (i.e. without using weakening) because in a way you have already supposed a sort of weakening in your hypotheses $(P \otimes Q) \multimap P$. And analogously to prove $(P \otimes Q) \multimap P$ from $P \multimap (Q \multimap P)$.

Remark for specialists: you can't prove the "exportation law" in linear logic if you replace the conjunction with the "with" $\&$. What linear logic shows is that the exportation law holds in a linear way in intuitionistic and classical logic if and only if you use the conjunction in a multiplicative way, it does not hold if you use the conjunction in an additive way. This is natural because the linear implication in linear logic is multiplicative.

Concerning relevance logic, someone added a link to the right answer in the comments below.

Answer 2

As per Taroccoesbrocco's answer, currying is already valid in linear logic. I think part of your confusion here might be that you are considering a restriction of the usual, more general currying schema $P \to (Q \to R) \equiv (P \wedge Q) \to R$, which becomes $P \multimap (Q \multimap R) \equiv (P \otimes Q) \multimap R$ in the notation of linear logic (and is closely related to the tensor-hom adjunction in monoidal closed categories). That is, you are considering the special case of the currying equivalence where $R = P$. In this case, the two sides of the equivalence both happen to be classical/intuitionistic tautologies, but that is irrelevant for the proof that the equivalence itself is a tautology (of classical/intuitionistic logic, but also of linear logic).