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- - **addition is well-defined**
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addition is well-defineddefine 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? |

Re: addition is well-definedOne 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. |

Re: addition is well-definedUnfortunately, induction will not work for this proof. It is a common mistake that I have seen tried a lot. |

Re: addition is well-definedLet 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. |

Re: addition is well-definedNot sure what you are trying to do. |

Re: addition is well-definedDman 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 h:omega-->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 Am:omega-->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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