As defined by Gan, a *dioperad* consists of sets of operations $P(n,m)$ with "$n$ inputs and $m$ outputs", which can be composed by joining *one* output of one operation to *one* input of another, giving rise to composition operations

$$ P(n_1,m_1) \times P(n_2,m_2) \to P(n_1+n_2-1,m_1+m_2-1). $$

Dioperads are apparently motivated by the fact that some PROPs are freely generated by dioperads, because they have a generators-and-relations description in which all the relations only involve composites of the sort that exist in a dioperad. In such a case, the dioperad is evidently a simpler object to study than the entire PROP.

As defined by Szabo, a polycategory has a set of objects, and for any $n$-tuple and $m$-tuple of objects $a_1,\dots,a_n$ and $b_1,\dots,b_m$, a set of morphisms $P(a_1,\dots,a_n; b_1,\dots,b_n)$, with composition operations that join one of the targets of one morphism to one of the sources of another. If we write $\Gamma,a,\Delta$ for a list of objects containing $a$ somewhere in the middle, then this composition operation looks like

$$P(\Gamma; \Delta,a,\Psi) \times P(\Xi,a,\Upsilon; \Phi) \to P(\Xi,\Gamma,\Upsilon; \Delta,\Phi,\Psi).$$

Polycategories are motivated by providing a semantics for classical linear logic, whose cut rule looks like this composition rule:

$$ \frac{\Gamma \vdash \Delta,A,\Psi \qquad \Xi,A,\Upsilon \vdash \Phi}{\Xi,\Gamma,\Upsilon \vdash \Delta,\Phi,\Psi} $$

One generally thinks of the source objects as joined by a "multiplicative conjunction" tensor product $\otimes$ and the target objects as joined instead by a "multiplicative disjunction" tensor product $⅋$, and this can be made precise by showing that "representable" polycategories are precisely linearly distributive categories.

It *looks* to me as though (ignoring questions of enrichment) a dioperad is just a one-object polycategory, in the same way that an ordinary operad is a one-object multicategory — or in other words, a polycategory is a "colored dioperad". Is this true? If so, is it in print anywhere? If not, what is the difference?