Reduced alternating diagrams minimize crossing number

If $D$ is empty, it has zero crossings. It is a regular diagram of $L$, so [F1] gives $0\le c(L)\le c(D)=0$, proving the claim in this case without evaluating a bracket. Henceforth $D$ and the represented link $L$ are nonempty.

Let $E$ be any regular diagram representing the same underlying unoriented link type as $L$. Since $L$ is nonempty, so is $E$. If its projection were disconnected, [F2] would make $L$ split, contradicting [F3] applied to $D$ and [F6]. Hence every competing diagram $E$ is connected. Orient each component of $E$ so an ambient isotopy to $L$ preserves its orientation.

By [F4], ${\mathrm{span}}_A⟨D⟩=4c(D)$ and ${\mathrm{span}}_A⟨E⟩\le 4c(E)$. The normalized bracket multiplies each raw bracket by a monomial, which shifts all exponents equally and preserves span. By [A1] and [F5], the normalized bracket values for the nonempty oriented diagrams $D$ and $E$ agree. Therefore $4c(D)\le 4c(E)$ and $c(D)\le c(E)$.

The diagram $D$ is one of the regular diagrams in the nonempty crossing-count set of [F1], so $c(L)\le c(D)$. Step 3.1 shows that $c(D)$ is no larger than any member of that set, and therefore $c(D)\le c(L)$. Combining the inequalities proves $c(D)=c(L)$.