January 4th, 2011, 11:14 PM  #1 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  addition is welldefined
define addition of natural number as follows: a) for every x, define x+1=x' b)for every x and y, defined x+y'=(x+y)' show that this addition is welldefined. That is, show that for all x and w, the value of x+w is welldefined. Hint: you may want to used induction. My question is what does it mean to be welldefined in this problem? 
January 5th, 2011, 01:13 AM  #2 
Senior Member Joined: Nov 2010 Posts: 502 Thanks: 0  Re: addition is welldefined
One would expect this operation to work on all natural numbers, and for every natural number to appear as a result of this operation. Further, only natural numbers should appear.

January 5th, 2011, 06:50 AM  #3 
Senior Member Joined: Dec 2008 Posts: 306 Thanks: 0  Re: addition is welldefined
Unfortunately, induction will not work for this proof. It is a common mistake that I have seen tried a lot.

January 5th, 2011, 11:49 AM  #4 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  Re: addition is welldefined
Let M be the set of x. can you set x=1 and w=y' ? so that x=1 => x+1=x'=1' and x+(y')= (y')'= (x+y)' this implies that 1 belong to M. 
January 5th, 2011, 07:39 PM  #5 
Senior Member Joined: Dec 2008 Posts: 306 Thanks: 0  Re: addition is welldefined
Not sure what you are trying to do.

January 10th, 2011, 11:51 AM  #6 
Senior Member Joined: Jan 2011 Posts: 106 Thanks: 0  Re: addition is welldefined
Dman is actually right. What you need here is recursion theorem on omega. In general, this theorem states that for any set A, an element a in A, and given f:A>A, there exist a unique hmega>A s.t h(0)=a h(n')= f(h(n)) for all n in omega. Then by recursion theorem we may define for all m in omega Ammega>omega Am(0)=m Am(n')= Am(n)' Then let +={((m,n),p) m and n are elements of omega ^ p= Am(n)} 

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