The BMW Markov trace gives the Dubrovnik link invariant

Fix an oriented link $K$. By [F1], choose a finite signed braid word ${\beta }_0$ whose oriented closure represents $K$. This use of existence is valid under [A1], as required by [F1]. Define $Q(\beta )$ by the formula in the statement for every braid word with closure equivalent to $K$.

A conjugation $\beta \mapsto \gamma \beta {\gamma }^{-1}$ leaves the closure component count unchanged because the strand permutation is conjugated, and leaves writhe unchanged because the exponent sum of $\gamma$ cancels with that of ${\gamma }^{-1}$. Cyclicity in [F3] gives ${\tau }_n(\gamma \beta {\gamma }^{-1})={\tau }_n(\beta )$. Thus $Q$ is unchanged by conjugation.

For positive stabilization ${\beta }^{\mathrm{\prime }}={\iota }_n(\beta ){\sigma }_n$, the new transposition involving strand $n+1$ joins that new strand to the cycle containing strand $n$, so $c({\beta }^{\mathrm{\prime }})=c(\beta )$. The exponent sum increases by one, hence $w({\beta }^{\mathrm{\prime }})=w(\beta )+1$. By [F3], ${\tau }_{n+1}({\beta }^{\mathrm{\prime }})=a{d}^{-1}{\tau }_n(\beta )$. Therefore $Q({\beta }^{\mathrm{\prime }})=-(-1{)}^c{a}^{-(w+1)}{d}^{n}a{d}^{-1}{\tau }_n(\beta )=Q(\beta )\mathrm{.}$ For negative stabilization the cycle count is again unchanged, the writhe becomes $w-1$, and ${\tau }_{n+1}(\beta {\sigma }_n^{-1})=a^{-1}{d}^{-1}{\tau }_n(\beta )$; substituting gives $-(-1{)}^c{a}^{-(w-1)}{d}^n{a}^{-1}{d}^{-1}{\tau }_n(\beta )=Q(\beta )\mathrm{.}$ Thus both stabilizations and their inverse moves preserve $Q$.

By [F2], any two braid words whose oriented closures represent $K$ are Markov equivalent. Each move in a finite Markov sequence preserves $Q$ by steps 2.1 and 2.2, so $Q$ is independent of the braid representative. For a representative $\beta$, [F7] identifies $\widehat{{\mathrm{\Phi }}_n(b_{\beta })}$ with its oriented closure diagram. By [F3], ${\tau }_n(\beta )=d^{-n}P(\widehat{{\mathrm{\Phi }}_n(b_{\beta })})$. Hence $d^{n-1}{\tau }_n(\beta )=d^{-1}P(\widehat{{\mathrm{\Phi }}_n(b_{\beta })})=D_{\widehat{\beta }}(a,z)$ by [F6]. Applying the conversion in [F6] gives $F_{\widehat{\beta }}(ia,-iz)=(ia{)}^{-w}{L}_{\widehat{\beta }}(ia,-iz)=-(-1{)}^c{a}^{-w}{D}_{\widehat{\beta }}(a,z)=Q(\beta )\mathrm{.}$ The scalar conversion lemma proves that $F$ is unchanged by each oriented Reidemeister move; [F8] therefore identifies this fixed-diagram value with the declared ambient-isotopy value of $K$. AC is used in [F1], [F2], and [F8]; the trace and local formula calculations are choice-free.