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The explanation:

1. Crossings in R^3 or S^3 (positive space)
2. Label 1+,1-,ect. The locations of D-joinings/crossings (mixed space)
3. Place a, b at each D.
4. Move to negative space: move crossing now a sharing as we are capable of having 2 parts or more at a vertex with a and b always moving along.

1+,1-,ect. Are the specific locations in R^3 where we put the other concept sharing diagram back together.

But in the moving diagram they have some freedom. They can travel along as they are labels, reachable through moves of the diagram (that is moves are reversible) or they can stop at a specific pair of locations and the a, b pairs move on forward. This is still reversible as I can get back to this D, as I can reverse from forward motion. At the end, after I go all the way back I come back to the same location.

So 1+ and 1- label the crossing 1, where the original locations can be traced back from. I am placing the labels on specific locations. I mean that this is where the crossing is so that I can trace back to the original location which matches moves in this space then are fully reversible and take full advantage of the new freedoms. Then all these diagrams are equivalent and complete as long as we don’t cut the diagram or change the order of locations.

Then R1,R2,R3, isotopy have their equivalents as well. I have rotation of the locations, creation of labels, joinings, movements of joinings and labels through sharings. And that’s all (complete).

Let there be another diagram D(2) in R^3 and we wish to compare this to the original diagram. Move it to mixed space and then to negative space. Concept share it and one part of it moves off. If we can make a congruency between this and the other diagram in negative space then the two diagrams in positive space are also congruent.

So we need to look for a match of the labels, joinings when we simplify the diagrams.

If they are the same, then the information should be contained in one diagram. That is, movements of one diagram should be able to produce the second one. So we need to look at one diagram to see if we can produce another(using all moves available)

Let’s look at T. To cut down a lot of the moves, let’s apply a simplifying method.

Now I only need to apply S1,S2,S3 and isotopy(2) and I can obtain any diagram of T. Then reform the diagram to go back to R^3.

But the only new diagrams I can obtain will be through the application of the new moves to the T-invariant. That is all possible moves of the invariant. None yield the unknot or any achirality.

The new moves are rotation (which isn’t usually considered) and movement of labels and joinings through sharings. 

So starting with the standard diagram for the Trefoil knot. I place this in the new space and I have the sharings D1,D2,D3. These have the associated moving parts a1,b1,a2,b2,a3,b3. Then I move the outer loops inward (green and red) forming the sharings Q1,Q2,Q3,Q4,Q5,Q6. These have the associated moving labels r1,s1,r2,s2,r3,s3. Then I form the joinings (a1(r1), (b2(s2), (a3(r3), (b1(s1), (a3(r2), (b3(s3) (all red). Now this can be further decomposed to eventually form the “circle”.

Showing the third move. After reduction, only Q-types are involved.

Here I show why we must keep to the same orientation when we reform a joining after we undo it in negative space.

Here is the achirality of the knot 4(1) shown in negative space.

Here is a better post on simplifications.

I place the trefoil, partially back together then move the four joinings through the reformed 2 and 3 sharings until they can possibly form a new diagram. I show it is not possible to form this diagram in positive space so that the trefoil is knotted.

From Knotscape I have found this diagram of Perko A. Crossings 1 through 10 shown (the orientation of the crossings written below the number, all are negative except number 9). Also I found a diagram of Perko B which is shown later. I proceed to move these diagrams into negative space and “undo” them sometimes adding Q-type sharings adding in s and r point(2)’s. This proceeds until I reach a diagram which has the Trefoil as a base for both diagrams. Then I rearrange the joinings to complete the congruency. This is all shown in detail below.

Sharing 9 is actually not reduced to (b9(a9) but can be moved to the outside of the trefoil base. There I can put it into congruence with the loop formed from (a17(b17) expressed as (b17(a17).