Another thing I "know" since 30 years but only recently I found a clue:          
Assume a R matrix (graphic: %) can be decomposed into "projectors" (graphic: >=< ).
(See the conditions here: http://mathoverflow.net/questions/65332/matrix-decomposition-the-other-way). It seems to me that this is always possible if it's the R matrix of
a Lie group, but *may* work even if not (i.e. just a random Yang-Baxter solution).
In any case, you can apply the usual trick - if you can pull a string over all nodes >= ,
i.e. Reidemeister 3 for nodes holds, Reidemeister 3 for crossings follows. Of course your
projectors must obey the usual 6j rules either. Now I found something fascinating:
The diagrams for Biedenharn-Elliott and Reidemeister 3 for nodes have five open ends.
I played with "quadratic" graphs with only four open ends, and the results are
exactly the same - the consistency rules following from the diagrams are exactly
Biedenharn-Elliott and Reidemeister 3 for nodes again.         
After "undecomposing" everything again, the following weakened Reidemeister 3 version
results: <IMG SRC="https://i.sstatic.net/gfOKy.jpg"> (hope the image comes through - https://i.sstatic.net/gfOKy.jpg)     
The above musings are not even a handwaving proof, so I won't dare to state "For Lie groups, Reidemeister 3 follows from weak Reidemeister 3". Still, it would be epic if one could replace the annoying 3+3-tangle condition with 2+2. And when I solve abstract tensor equations, I know not a single counterexample either. So my question: Do you know a
counterexample - any construction where weak Reidemeister 3 holds but Reidemeister 3 is violated?