Proof
The ≤ direction is obvious: we can write a program to produce x and y by concatenating a program to produce x, a program to produce y given access to x, and (whence the log term) the length of one of the programs, so that we know where to separate the two programs for x and y|x (log(K(x, y)) upper-bounds this length).
For the ≥ direction, it suffices to show that for all k,l such that k+l = K(x,y) we have that either
K(x|k,l) ≤ k + O(1) or K(y|x,k,l) ≤ l + O(1).Consider the list (a1,b1), (a2,b2), ..., (ae,be) of all pairs (a,b) produced by programs of length exactly K(x,y) . Note that this list
- contains the pair (x,y),
- can be enumerated given k and l (by running all programs of length K(x,y) in parallel),
- has at most 2K(x,y) elements (because there are at most 2n programs of length n).
First, suppose that x appears less than 2l times as first element. We can specify y given x,k,l by enumerating (a1,b1), (a2,b2), ... and then selecting (x,y) in the sub-list of pairs (x,b). By assumption, the index of (x,y) in this sub-list is less than 2l and hence, there is a program for y given x,k,l of length l + O(1). Now, suppose that x appears at least 2l times as first element. This can happen for at most 2K(x,y)-l = 2k different strings. These strings can be enumerated given k,l and hence x can be specified by its index in this enumeration. The corresponding program for x has size k + O(1). Theorem proved.
Read more about this topic: Chain Rule For Kolmogorov Complexity
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