TITLE: What is the intuition behind the "par" operator in linear logic?
QUESTION [28 upvotes]: I'm $\DeclareMathOperator{\par}{\unicode{8523}}$ trying to wrap my mind around the $\par$ ("par") operator of linear logic. 
The other connectives have simple resource interpretations ($A\otimes B$ is "you have both $A$ and $B$", $A\&B$ is "you can have either $A$ or $B$" etc.).
But despite the inference rules being simple, I've seen only mentions that $A\par B$ is "difficult to describe" in a resource interpretation, apart from the special case of $A\multimap B ~\equiv ~ A^\bot \par B$ ("you can convert $A$ to $B$").
Any way I can understand it better?

REPLY [4 votes]: $\DeclareMathOperator{\par}{⅋}$I found it helpful to generalize the resource analogy to deferred promises of resources, similar to the notion of "promises" or "futures" in asynchronous programming. Thus, $\mathit{Apple}$ is not necessarily a real apple, but could be a promise of an apple from someone else.
This distinction makes it possible to interpret the negation $\mathit{Apple}^\bot$ as an obligation to surrender an apple to someone else.
In general, a proposition $A$ in linear logic may contain a mixture of promises and obligations. So something like
$\mathit{Pie}^\bot \otimes \mathit{Apple}$
represents the obligation to provide a pie and, alongside, the promise of an apple.
Distinction between $\par$ and $\otimes$
The interpretation of par ($\par$) is intimately related to the duality of promises and obligations. For example, one may interpret
$\mathit{Flour}^\bot \par \mathit{Cake}$
as the obligation to provide flour with the promise of cake, such that they may be contigent on one another. The chef might require the flour to bake the cake, so you cannot expect to have the cake before flour has been given.
The presence of contingency in the fulfillment of these promises or obligations is what sets $\par$ apart from $\otimes$.

With $A \otimes B$, you have set of promises and obligations $A$, and a set of promises and obligations $B$, and neither set is contingent on the other. This means $A$ and $B$ must have been constructed through totally independent mechanisms.

With $A \par B$, the fulfillment of promises in $A$ may be contingent on certain obligations in $B$ and vice versa. From a software perspective, one could think of $A$ and $B$ as independent threads of execution, with obligations acting as the sending end of channels and promises as the receiving end.


Dually, to prevent cycles of mutually dependent promises, which could cause a deadlock, operations on $A \par B$ must be disjoint: they are limited to those that can act on $A$ independently of $B$ and vice versa. For example, you can transform $A$ into $C$ and $B$ into $D$:
$$A \par B, A \multimap C, B \multimap D \vdash C \par D$$
However, you cannot in general act on both $A$ and $B$ at the same time:
$$A \par B, A \otimes B \multimap C \nvdash C$$
Introduction
As noted previously, $\par$ closely related to obligations (negation). In fact, the only way to create nontrivial $\par$ out of thin air is by creating an obligation:
$$\vdash A^\bot \par A$$
Once created, you are free to transform either side of $\par$ as long as you do them disjointly of the other side.
Note that the contingency does not have to be one-sided; it is totally possible to have mutual contingency as long as there is no risk of cycles:
$$\vdash (\mathit{Flour}^\bot \par \mathit{Cake}) \par (\mathit{Flour} \otimes \mathit{Cake}^\bot)$$
Elimination
To get rid of a $\par$, you have to fulfill all the obligations (and use up all the promises) on either the left or right side of $\par$:
$$A \par B, B \multimap \bot \vdash A$$
Here, $\bot$ is the identity element of $\par$ and signifies the state of having no obligations remaining so you can slack off indefinitely. It can be created by connecting a promise with an obligation:
$$A, A^\bot \vdash \bot$$
Contrast with $\mathbf{1}$, which signifies a promise of nothing, but you still need to continue working to fulfill your obligations.