Given a complete partial order $(D,\, {\leq})$ and a Scott-continuous function $F\colon D \rightarrow D$ and some fixed point $X = F(X)$.

Are there any criteria that ensure that $X = \mathsf{lfp} F$ where $\mathsf{lfp} F$ is the least fixed point of $F$?

Furthermore, are there any criteria that ensure that $Y \leq \mathsf{lfp} F$ for an arbitrary point $Y$?

  • 1
    You mean without computing the LFP? To check if y is less than LFP you can check if for some n $y < F^n(\bot)$ which seems only semidecodable in general. – Kaveh Nov 12 '15 at 17:01
  • Yes, I mean without computing or knowing the LFP. I'm looking for something more elegant than finding such an $n$. The method need not be complete, just sound. – zander Nov 12 '15 at 18:00

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