= Solution
Put $\delta=-A^2-A^{-2}$. The reduced <Kauffman bracket> is the <Laurent polynomial> determined by $\langle\bigcirc\rangle=1$, $\langle D\sqcup\bigcirc\rangle=\delta\langle D\rangle$, and the crossing expansion $\langle D\rangle=A\langle D_A\rangle+A^{-1}\langle D_B\rangle$. Fix the A smoothing so that a positive curl has multiplier $-A^3$; rotating the local crossing interchanges the smoothing descriptions. Equivalently, its <bracket smoothing states> give
$$
\boxed{\langle D\rangle=\sum_sA^{a(s)-b(s)}\delta^{|s|-1},}
$$
where $a(s)$ and $b(s)$ count its A and B smoothings and $|s|$ is the number of state <circles>. Resolving all crossings proves that the recursive rules define the same <polynomial> independently of the order of expansion.
Under a positive <Reidemeister move> of type I, the two resolutions give $A\delta+A^{-1}=-A^3$ times the straight strand; the negative curl gives $A^{-1}\delta+A=-A^{-3}$. Thus the bracket itself is not an unframed <link invariant>. For type II, use the adjacent cap-cup generator $e$ of the <Temperley-Lieb algebra>, with $e^2=\delta e$. The crossing and its inverse are $R=A1+A^{-1}e$ and $R^{-1}=A^{-1}1+Ae$, and
$$
RR^{-1}=1+(A^2+A^{-2}+\delta)e=1.
$$
For type III, $e_ie_{i+1}e_i=e_i$ and the analogous reversed relation give
$$
R_iR_{i+1}R_i-R_{i+1}R_iR_{i+1}=A^{-1}(A^2+\delta+A^{-2})(e_i-e_{i+1})=0.
$$
Hence the bracket is invariant under types II and III.
For an oriented diagram, let $w(D)$ be its <writhe of a link diagram>, the sum of its signed crossings. Types II and III preserve <writhe of a link diagram>, while a positive or negative curl changes it by one or minus one. The <Jones polynomial> is therefore
$$
\boxed{V_L(t)=(-A^3)^{-w(D)}\langle D\rangle,\qquad t=A^{-4},\qquad V_{\bigcirc}=1.}
$$
The normalization cancels type I, so all <Reidemeister moves> preserve it. It is a <Laurent polynomial> in $t^{1/2}$; its breadth means the largest occurring t exponent minus the smallest, including half-integral exponents for <links>.
The <Jones polynomial skein relation> follows directly from the same normalization. Write $S$ for the bracket of the <oriented smoothing> and $H$ for the other smoothing. The local expansions are $\langle D_+\rangle=AS+A^{-1}H$ and $\langle D_-\rangle=A^{-1}S+AH$, while $w(D_\pm)=w(D_0)\pm1$. Consequently
$$
A^4V_{L_+}-A^{-4}V_{L_-}=(-A^3)^{-w(D_0)}(A^{-2}-A^2)S.
$$
Substituting $t=A^{-4}$ gives
$$
\boxed{t^{-1}V_{L_+}-tV_{L_-}+(t^{-1/2}-t^{1/2})V_{L_0}=0.}
$$
The other smoothing has cancelled, so the identity applies with exactly the oriented local convention in the PDF.
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