Talk:Recursive definition

message: Isn't this subject already covered by Recursion, though?  -- Antaeus Feldspar 18:41, 7 Oct 2004 (UTC)
replies: text: I wondered that, too, but then I read the recursion article. My article can stand alone, although I wouldn't mind if someone merged the two together. --Uncle Ed 19:08, 7 Oct 2004 (UTC)

text: These seems to be recursive definitions not recursive functions.  Is that a distinction?  RJFJR (talk) 16:52, 14 July 2009 (UTC)
replies: text:  Yes, so I oppose the merge. These are distinct notions. An important application in type theory (and thus type systems) is the recursive data type. Pcap ping 07:56, 18 August 2009 (UTC)

text:  Oppose the merge. This article stands by itself; it has multiple uses, one of which is as a sub-page of definition. --Ancheta Wis (talk) 12:16, 19 August 2009 (UTC)
thread_title:  merge 

message: removed because as far as I can tell these are circular rather than recursive definitions, since they lack a base case. RJFJR (talk) 16:52, 14 July 2009 (UTC)
thread_title: Recursive definitions in hacker culture

message: Shouldn't the letters be replaced by the actual logic symbols? Dude1818 (talk) 00:11, 21 December 2009 (UTC)
thread_title:  WFFs 

message: The proof that recursive definitions are legit needs an overhaul. Right off the bat, the initial supposition that one has both the functions f and g begs the whole question. The function f is the very one whose existence is supposed to be justified by the theorem. What you want to prove (in the simplest case) is that, given ONLY a function g and an object b, there exists a function f on ℕ satisfying f(0) = b and f(n+1) = g(f(n)), for all n∈ℕ. The proof proceeds in two parts. First, by induction (or, equivalently, the least number principle), one shows that, for every n∈ℕ, there is a (finite) function Fn on {0,...,n} such that Fn(0) = b and, for m<n, Fn(m+1)=g(Fn(m)). (The current proof gestures in this direction but the effort is vitiated by the references to the function f whose existence is to be proved.) 
In the second part of the proof, one defines the desired function f (now called "F" in the proof) to be "the set theoretic union of all the Fn". However, you can't just take the union of "all" of a bunch of things, as the current proof would have it. You first have to prove that they exist together as the members of a set. That is proved in this case as follows: Every Fn is a subset of ℕxℕ and hence is a member of the powerset ℘(ℕxℕ) of ℕxℕ. Hence, the set M of all of the Fn exists by the axiom of Separation: M = {h ∈ ℘(ℕxℕ) : h is Fn, for some n}. The desired function f given by the recursion can now be proved to exist by the Union axiom: f = ∪M.  
I don't have time to gussy this up for the entry at the moment, but it needs to be done. I'll do it eventually if no one else does. TXlogic (talk) 14:40, 24 November 2013 (UTC)
thread_title:  Problems in the proof of the definition by recursion theorem  

message: "Note that this definition assumes that N is contained in a larger set (such as the set of real numbers) — in which the operation + is defined." is not correct. 31.49.9.229 (talk) 22:19, 29 December 2019 (UTC)
thread_title:  Error