On Mar 9, 5:57 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> What do you understand by the not changeable function (d)?
If x is a potentially infinite sequence of 0's and 1's then we say (x) is associated to x, if (x) is an algorithm which given a natural number produces a finite string of 0's and 1's, such that for every natural number n, (x) produces the nth FIS of x.
We will say x is coFIS to (y) iff
i. We have (x) associated to x and (y) associated to y
ii. For every n, (x) and (y) produce the same finite string.
Some comments.
We can never have more than a limited number of the strings produced by (x) existing at any time, since we can never have more that a limited number of natural numbers.
We cannot show that x is coFIS to (y) by using (x) and (y) to produce "all possible strings" and comparing them, since "all possible natural numbers" does not exist. However, we may be able to use induction to show "For every n, (x) and (y) produce the same finite string"
To show that x is coFIS to (y), it is not enough to show that every existing FIS of x, is equal to an existing FIS of y.
d, the diagonal of the list L, is a potentially infinite sequence of 0's and 1's with associated (d): for any natural number n produce a sequence of n ones.
l_k, the k_th line of L is a potentially infinite sequence of 0's and 1' with associated (l_k): for any natural number n produce a sequence of k 1's followed by a sequence of n-k 0's
A line of L, l_k is findable iff the index k is findable.
Do you agree with the statement
There does not exist (in the sense of not findable) a natural number m such that the mth line of L is coFIS to (d)
