Re: Weird problem


Bill Taylor wrote:

> f(x) = 1 if CH is true; 2 if it is false.
>
> This one seems to be different; because whereas FLT was entirely
> expressible in PA, CH is not so.

Every constant function is recursive. Thus if the above defines
a function, it is recursive. Whether or not the above defines
a function is a matter of opinion - in standard set theory it
certainly does. On the other hand, a definition such as

f(x) = 1 if Bill Taylor is an irritating git; 2 if he is not

does not define a function in standard mathematics, and so it
makes no mathematical sense to say "f is primitive recursive"
or "f is not primitive recursive".

I hope this answers your question, you redoubtable old
sheep-shagger! (A designation I first had occasion to
use when you bested Ilias Kastanas.)

.