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I define a fuction $*$ in a recursive way on a subset of natural numbers (with subset I mean that is $*$ define in the second argument only for $\Bbb N-\{0,1\}$ in other words we have $*:\Bbb N \times (\Bbb N-\{0,1\})\rightarrow \Bbb N$

$$m * 2:=m+m$$ $$m* S(n+2):=m+(m*(n+2))$$

I'm not sure if a proof of the non-totality of $*$ on the natural numbers is needed:

$A$-The non-totality is, in my opinion, self-evident and doesn't need a proof because the definition defines it only for $\Bbb N-\{0,1\}$

$B$-but maybe it needs a proof because I'm stating that a property $\phi$ holds for the function I've defined (exist a number $n_0$ such that for every pair of the form $(n_0,m)$ we have that $(n_0,m)\notin f$ is true)

So my question is:

Do I need to give a proof even if the non totality is self-evident from the recursive definition? If I have to give a proof, what can be its "shape" ?

MphLee
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  • As a good rule of thumb, omit the proof of a theorem iff that proof is straightforward. So in particular, if a theorem is "obvious" but does not have an easy proof, you need to include a proof. See also, obvious theorems that are false. – goblin GONE Jun 05 '14 at 10:51
  • @user18921 yes I agree but I don't know how to apply this rule for the totality of $\cdot'$: I don't know if I need to show the proof because I don't even know it exist (if the proof is the recursive definition itself) – MphLee Jun 05 '14 at 11:55
  • For the record, I do think its obvious, but I don't know how to prove it. In general, it seems to be very hard to find information about how to use recursive definitions "rigorously." So perhaps that is what you should look for; a book or article that gives a rigorous account of recursion. – goblin GONE Jun 05 '14 at 12:03
  • By the way, your notation is confusing look. I'd replace $\cdot'$ with $*$ or some other symbol. – goblin GONE Jun 05 '14 at 12:58
  • I'm not clear on what susbset you have it defined on and what it's defined as. Rewriting needed before answers will come. – Daniel Donnelly Jun 05 '14 at 15:56
  • @EnjoysMath $*$ is defined only if the second argument is greater than $1 $ – MphLee Jun 05 '14 at 16:38

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