3 ms·
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
by j-pb 2y ago
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.