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I'm basically trying to solve 4.2 enter image description here

(Taken from Semantics with Applications)

As I see it, the functional F will be defined like so:

(F g)s = s if s x == 0 
(F g)s = gs[x -> x -1] if s x!=0 

So, the first function is

g1 s = undefined for all s 

I don't think this is a fixed point, because when s x = 0 , the function would output undefined when it really should be s

Am I on the right track?

By that logic, g2 and g3 would be fixed points since they terminate at a state where x = 0.

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    $\begingroup$ Don't use images as main content of your post. This makes your question impossible to search and inaccessible to the visually impaired; we don't like that. Please transcribe text and mathematics (note that you can use LaTeX) and don't forget to give proper attribution to your sources! $\endgroup$ – dkaeae Apr 17 at 11:51
  • $\begingroup$ Do you have a more specific question? "Am I on the right track?" might not be the most useful to inspire answers that will be helpful to others in the future. We're happy to help you understand the concepts but just solving exercises for you is unlikely to achieve that. You might find this page helpful in improving your question. Are you able to identify some conceptual issue that you're confused on? Can you ask about that instead of asking us to check your work on this exercise or to solve this exercise for you? $\endgroup$ – D.W. Apr 17 at 15:56

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