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January 4th, 2011, 11:14 PM   #1
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addition is well-defined

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 well-defined. That is, show that for all x and w, the value of x+w is well-defined. Hint: you may want to used induction.

My question is what does it mean to be well-defined in this problem?
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January 5th, 2011, 01:13 AM   #2
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Re: addition is well-defined

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.
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January 5th, 2011, 06:50 AM   #3
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Re: addition is well-defined

Unfortunately, induction will not work for this proof. It is a common mistake that I have seen tried a lot.
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January 5th, 2011, 11:49 AM   #4
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Re: addition is well-defined

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.
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January 5th, 2011, 07:39 PM   #5
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Re: addition is well-defined

Not sure what you are trying to do.
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January 10th, 2011, 11:51 AM   #6
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Re: addition is well-defined

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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