[sbazerque]

Did you read the remark at the end of my comment? In the practical cases I was exploring, that combinatorial explosion does not happen. It's relaxed in the sense that it is coordination-free.

[j-pb]

Not sure what you mean.

I'm talking about the "relaxed" P' being defined via the power set of S.

2^S= {s | s ⊆ S}

Now if all your P is only a mapping then

P'(S) = {<s, P(s)> | s ∈ S}

but then your "coordination free" P was monotonic anyways.