= Solution
\b[True with the stated Conway normalization.] Fix $t^{1/2}=i$ at $t=-1$. The <Jones polynomial skein relation> becomes
$$
V_{L_+}(-1)-V_{L_-}(-1)=-2iV_{L_0}(-1).
$$
The <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 $W_{L_+}-W_{L_-}=-2iW_{L_0}$. 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> gives
$$
\boxed{V_K(-1)=\nabla_K(2i)=\Delta_K(-1).}
$$
This is the <signed Jones evaluation at minus one>. Replacing $\Delta_K$ by an arbitrary <Laurent unit> multiple would destroy the signed conclusion.
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