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The Conway polynomial is normalized by ${\nabla }_U=1$ for the unknot, zero for a split unlink with more than one component, and the skein relation ${\nabla }_{L_{+}}-{\nabla }_{L_{-}}=z{\nabla }_{L_0}$. For a knot, the symmetric Alexander polynomial with value one at $t=1$ is ${\Delta }_K(t)={\nabla }_K(t^{1/2}-t^{-1/2})$. This fixes the Laurent-unit ambiguity of the Seifert-matrix presentation of the Alexander module.

Resolve the two crossings in a skein relation using the Kauffman bracket, then correct by the writhe of a link diagram. The other smoothing cancels, leaving the displayed coefficient of the oriented smoothing. This relation and the value one on the unknot determine the Jones polynomial: crossing changes give descending diagrams, which are unlinks, while the smoothing terms have fewer crossings.

The oriented smoothing replaces a crossing by two disjoint arcs whose orientations agree with the four oriented ends. It is the zero-resolution in an oriented skein relation.

True with the stated Conway normalization. Fix $t^{1/2}=i$ at $t=-1$. The Jones polynomial skein relation becomesThe Conway polynomial satisfies ${\nabla }_{L_{+}}(z)-{\nabla }_{L_{-}}(z)=z{\nabla }_{L_0}(z)$. Put $W_L=(-1{)}^{\ell (L)-1}{\nabla }_L(2i)$. An oriented smoothing changes the number of components by one, so these quantities satisfy precisely $W_{L_{+}}-W_{L_{-}}=-2i{W}_{L_0}$. Both $W$ and $V(-1)$ are one on the unknot and zero on every multi-component unlink: for $V$, the split-union factor vanishes at this substitution.

These skein relations and unlink values determine the invariants uniquely. Switching a first undesired crossing gives a diagram closer to a descending diagram, while the smoothing term has fewer crossings; induction first on crossing count and then on undesired crossings reduces everything to unlinks. Thus $W_L=V_L(-1)$. For a knot there is one component and the Conway-normalized Alexander polynomial givesThis is the signed Jones evaluation at minus one. Replacing ${\Delta }_K$ by an arbitrary Laurent unit multiple would destroy the signed conclusion.

Apply the given skein relation to a positive and a negative kink in a diagram of the unknot. Both crossed link diagrams represent the unknot, whereas the oriented smoothing represents the two-component unlink. ThusFactor the left-hand side and cancel the nonzero factor in the Laurent polynomial ring:The positive sign follows from the skein relation as printed in this paper. It differs from the frequently used Jones polynomial convention whose two-component unlink has value $-(q^{1/2}+q^{-1/2})$. With the printed convention, an $r$-component link has invariant $(-1{)}^{r-1}$ times that convention's Jones polynomial.

Insert a kink on any component of an oriented link diagram for $L$. Switching the kink's crossing gives another link diagram of $L$; its oriented smoothing separates off a small unknot in a ball disjoint from the rest of $L$. The skein relation therefore givesCancel the nonzero factor $q^{1/2}-q^{-1/2}$ in the Laurent polynomial ring to obtain

Reflection through the projection plane reverses the orientation of $S^3$ and exchanges the overpassing and underpassing strands at every crossing. Thus switching all crossings gives the mirror of a link, with the component orientations transported by the reflection.

Define $W(L)(q)=V(\overline{L})(q^{-1})$. Reflection exchanges $D_{-}$ with $D_{+}$ and preserves the oriented smoothing. Apply the printed skein relation to the reflected triple at parameter $q^{-1}$:Multiplication by $-1$ gives the original skein relation for $W$. Also $W(U_1)=1$. The stated uniqueness of the Jones polynomial now gives $W=V$, hence

Let $H_{+}$ be the positively oriented Hopf link obtained by closing a two-strand braid with two positive crossings. Switching one crossing gives the two-component unlink; the oriented smoothing gives the unknot. The printed skein relation givesRepresent the right-handed trefoil knot by the closure of the two-strand braid with three positive crossings. Switching one crossing gives the unknot, while its oriented smoothing gives $H_{+}$. Consequentlyso the two Jones polynomials areThe second expression follows from the mirror of a link identity. They are unequal Laurent polynomials, so invariance under link isotopy proves that the right-handed and left-handed trefoils are not isotopic.