This is much like an argument seen over on Formal vs. Magical Usage. Let's start by recognizing that the statement

" Person 1 and Person 2 are the same person. "

can be viewed as a pair of statements :

" All who are Person 1, are Person 2" ....... and
" All who are Person 2, are Person 1"

Recognizing this, we can now set up a series of sorites sound enough to please those who wish to play Devil's advocate on this painfully obvious point. Let P(k) be the person who results after k cells have changed in your brain, for each integer (whole number) k; let N be the number of cells in your brain. Then the original you is P(0), P(N) is the carbon copy of the other person you are being transformed into, and our assumption that a single cell change doesn't change who you are becomes the assertion that P(k+1) is the same person as P(k), ie.

" All who are P(k) are P(k+1) " ....... and
" All who are P(k+1) are P(k) "

With these pairs of statements in mind, we can now set up the pair of sorites, in an obvious fashion :

and conversely, since "all who are P(k+1) are P(k)" for each k from 0 to N-1 is equivalent to "all who are P(m) are P(m-1)" for each m from 1 to N (just define m to be k+1, in each case)

These two conclusions, taken together, force one to conclude that your old self, who you are before the first cell replacement, is one and the same as your new self, the person who is created after all of the cells have been changed, since the old you, being one of those who is the old you, is one of those who is the new you, and vice versa. If this all seemed to be an exercise in hand-waving, no problem - we can easily make it into an exercise in Mathematical induction, instead, and wouldn't you have been so glad to have raised the point? For those who are unfamiliar with the concept, let us explain.

Given any nonempty set of non-negative integers (what your grade school teacher told you to call "whole numbers") must have a least member, one may argue as follows : Suppose, for the sake of contradiction, that while our assumption that

" All who are P(k) are P(k+1) "

held for all k from 0 to N-1, that the conclusion that

" All who are P(0) are P(N) "

was invalid. Then the set S, consisting of those integers r greater than or equal to 0, and less than or equal to N, such that the statement

" All who are P(0) are P(r) "

is false would be nonempty, as it would have at least one member : N. It would, therefore, have to have a least element, which we may call M. M certainly can not be 0; it is a tautology that all who are P(0) are P(0); therefore, we may write, without making use of a statement which we are not given, that

" All who P(0) are P(M-1) "

since M-1, being less than M (the least member of S) can't be a member of S, itself; and, taking k = M-1, since (M-1) + 1 = M, we have

" All who are P(M-1) are P(M) "

leaving us with the syllogism

" All who are P(0) are P(M-1) "
" All who are P(M-1) are P(M) "

Therefore : "All who are P(0) are P(M) "

which is a contradiction, since M is a member of S. Therefore, we must reject the assumption that the statement "All who are P(0) are P(N)" is false, given our starting assumptions, as it leads to an inescapable contradiction. Conversely, suppose that the statement

" All who are P(M) are P(0) "

was false. Then the set T, here defined as being the set of all integers q greater than or equal to 0, and less than or equal to N, such that the statement

" All who are P(N) are P(N-q) "

is false would, once again, be nonempty, having at least one member : N. As was the cas with S in the last argument, T would have to have a least member : let's call it L. As L-1, being less than L, would not be a member of T, we would be left with the syllogism

" All who are P(N) are P(N-(L-1)) "
" All who are P(N-(L-1)) are P(N-L) "

Therefore : "All who are P(0) are P(N-L) "

producing a contradiction much like before, forcing us to conclude that is empty, and that "all who are P(N) are P(0)". (Note : the second statement in the above syllogism is simply a restatement of one of our starting assumptions, since

N .. - .. (L .. - .. 1) .. = .. (N .. - .. L) .. + .. 1

meaning that it is simply the statement "all who are P(k+1) are P(k)" with k being taken to be N-L. Since k lies between 0 and N-1 if and only if N-k lies between 1 and N (I'll leave this easy proof to the reader), this is a legitimate substitution). This, again, presents us with the pair of identifications we saw above : "All who are your old self, are your new self", and vice versa, forcing the conclusion in question.

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