A simple question about Jones polynomial

Question

I understand the proof that the Jones polynomial, if its unique, is invariant under reidemeister moves, however how do we show that two inequivalent ways of computing it using the three rules and reducing it to unknots gives the same polynomial?

Answer 1

If you look at the formulation of the Jones polynomial as a sum over states, then it is clear from the start that the order in which you resolve crossings doesn't matter, since you are summing over all states and addition is commutative.

See here for more info on the state model of the Jones polynomial.