The choice of the steps in the proposed derivation and the stratification of the worlds represents an attempt to maximally avoid the resulting overlappings of the worlds and duplications. Certainly, after the derivation proposed, several natural questions appear: how can we be sure that the derivation proposed is exhaustive, and, moreover, how can we be sure that no duplications remain. The answer to the first question can be found in the graph-theoretical approach used by A. Caudron (1982). To solve the problem of possible duplications we can use graph-transformation methods developed by the same author, but also the program LinKnot and its function SameAltConKL (webMathematica SameAltConKL) that compares two alternating KLs and determines their equality. For example, for the two source links ((2,2),2),(2,2),2 and ((2,2),2) ((2,2),2) obtained at the end of our previous derivation the result is: yes, they are equal. For a future computer derivation of alternating KLs in Conway notation we can use an exhaustive algorithm for their derivation that produces a minimal number of duplications, and then apply to the KLs obtained the function SameAltConKL in order to eliminate any remaining duplications.

With regard to non-alternating KLs given in Conway notation, the situation is somewhat different. In principle, for all non-polyhedral KLs it is possible to construct derivation algorithms (similar to that used for the derivation of non-alternating KLs of the stellar world) that will produce all KLs (i.e., that will be exhaustive), and that will produce a minimum number of duplications.

After exhausting symmetry arguments, the tools we have for eliminating duplications are the KL reduction function KnotLinkReduce and functions for computing polynomial invariants. Even after their use, we cannot be sure that all duplications are eliminated.

From the alternating KLs belonging to the arborescent world for n = 8 we obtain the following non-alternating KLs:

 n = 8 (2,2) (2,2-) (2,2) -(2,2) No. of KLs: 2

 n = 9 (3,2-) (2,2) (3,2) (2,2-) (2 1,2-) (2,2) (2 1,2) (2,2-) (2,2+) (2,2-) (2,2+) -(2,2) No. of KLs: 6

 n = 10 (4,2) (2,2-) (2,2) (4,2-) (4,2) -(2,2) (3,3) (2,2-) (3,3-) (2,2) (3,2 1) (2,2-) (2 1,3-) (2,2) (2 1,2 1) (2,2-) (2 1,2 1-) (2,2) (3,2) (2 1,2-) (2 1,2) (3,2-) (2 2,2) (2,2-) (2 2,2-) (2,2) (2 1 1,2) (2,2-) (2 1 1,2-) (2,2) (3,2) (3,2-) (2 1,2) (2 1,2-) (3,3) -(2,2) (3,2 1) -(2,2) (2 1,2 1) -(2,2) (3,2) -(2 1,2) (3 1,2) -(2,2) (2 2,2) -(2,2) (2 1 1,2) -(2,2) (3,2) -(3,2) (2 1,2) -(2 1,2) (2,2,2) (2,2-) (2,2,2-) (2,2) (2,2,2-) (2,2-) (2,2,2--) (2,2) (2,2,2) -(2,2) (2,2) 2 (2,2-) (2,2) -2 (2,2) (2,2) 2 -(2,2) (2,2),2,(2,2-) (2,2),-2,(2,2) (2,2),2,-(2,2) (3,2+) (2,2-) (2 1,2+) (2,2-) (2,2+) (3,2-) (2,2+) (2 1,2-) (3,2+) -(2,2) (2 1,2+) -(2,2) (2,2+) -(3,2) (2,2+) -(2 1,2) (2,2++) (2,2-) (2,2++) -(2,2) (2,2,2) (2,2-) No. of KLs: 48

In order to write them as KLs with a minimum number of crossings, we can write -2 instead of 2-; -2,-2 instead of 2,2- -, etc.