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by j-pb
646 days ago
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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. |
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