Let us assume that
\({\mathbf {v}}_{A}=\psi (b_{1},b_{2},b_{3},b_{4})\) and that the label of point
\(D\) differs in two bits from that of
\(A\), i.e.,
\({\mathbf {v}}_{D}=\psi (\overline{b_{1}},\overline{b_{2}},b_{3},b_{4})\) (vertical labeling orientation). At first, let us decide on which labels can be assigned to points
\(M\) and
\(P\). According to Theorems
1 and
2, all the labels differing in one bit from
\({\mathbf {v}}_{A}\) and from
\({\mathbf {v}}_{D}\) are forbidden. Additionally,
\({\mathbf {v}}_{M}\) must not differ in four bits from
\({\mathbf {v}}_{A}\), and
\({\mathbf {v}}_{P}\) must not differ in four bits from
\({\mathbf {v}}_{D}\). Therefore,
$$\begin{aligned} {\mathcal {V}}_{M}&= \left\{ \psi \left( b_{1},\overline{b_{2}},b_{3},\overline{b_{4}}\right) \!,\psi \left( b_{1},\overline{b_{2}},\overline{b_{3}},b_{4}\right) \!,\psi \left( b_{1},\overline{b_{2}},\overline{b_{3}},\overline{b_{4}}\right) \!,\psi \left( \overline{b_{1}},b_{2},\overline{b_{3}},\overline{b_{4}}\right) ,\right. \nonumber \\&\quad \left. \psi \left( \overline{b_{1}},b_{2},b_{3},\overline{b_{4}}\right) ,\psi \left( \overline{b_{1}},b_{2},\overline{b_{3}},b_{4}\right) ,\psi \left( b_{1},b_{2},\overline{b_{3}},\overline{b_{4}}\right) \right\} \end{aligned}$$
(23)
and
$$\begin{aligned} {\mathcal {V}}_{P}&= \left\{ \psi \left( b_{1},\overline{b_{2}},b_{3},\overline{b_{4}}\right) \!,\psi \left( b_{1},\overline{b_{2}},\overline{b_{3}},b_{4}\right) \!,\psi \left( b_{1},\overline{b_{2}},\overline{b_{3}},\overline{b_{4}}\right) \!,\psi \left( \overline{b_{1}},b_{2},\overline{b_{3}},\overline{b_{4}}\right) \right. \nonumber \\&\quad \left. \psi \left( \overline{b_{1}},b_{2},b_{3},\overline{b_{4}}\right) ,\psi \left( \overline{b_{1}},b_{2},\overline{b_{3}},b_{4}\right) ,\psi \left( \overline{b_{1}},\overline{b_{2}},\overline{b_{3}},\overline{b_{4}}\right) \right\} \end{aligned}$$
(24)
It can be easily noticed that
\({\mathcal {V}}_{M}\) and
\({\mathcal {V}}_{P}\) are identical with the exception for their last elements. The labels of points
\(K,L,O\) must differ in one bit from
\({\mathbf {v}}_{A}\) since in the spectrum of
\(\alpha \) points there are single
\(8d\), and double
\(13d\) entries. To reinforce that clause we note that for point
\(A\) the only possible
\(8d\) distance is that to point
\(K\) and
\(13d\)—to points
\(L\) and
\(O\). Additionally,
\({\mathbf {v}}_{L}\) and
\({\mathbf {v}}_{O}\) cannot differ in one bit from
\({\mathbf {v}}_{D}\) since there are neither
\(4d\) nor
\(10d\) distances in the spectrum of
\(\alpha \) points. Therefore,
$$\begin{aligned} {\mathcal {V}}_{LO}=\left\{ \psi \left( b_{1},b_{2},\overline{b_{3}},b_{4}\right) ,\psi \left( b_{1},b_{2},b_{3},\overline{b_{4}}\right) \right\} \end{aligned}$$
(25)
and
$$\begin{aligned} {\mathcal {V}}_{K}&= \left\{ \psi \left( \overline{b_{1}},b_{2},b_{3},b_{4}\right) ,\psi \left( b_{1},\overline{b_{2}},b_{3},b_{4}\right) \right. \nonumber \\&\quad \left. \psi \left( b_{1},b_{2},\overline{b_{3}},b_{4}\right) ,\psi \left( b_{1},b_{2},b_{3},\overline{b_{4}}\right) \right\} . \end{aligned}$$
(26)
Since
\({\mathcal {V}}_{LO}\subset {\mathcal {V}}_{K}\) consists of exactly 2 elements, each of them must be assigned to either point
\(L\) or point
\(O\) and, therefore, cannot be assigned to point
\(K\). Thus we reformulate
$$\begin{aligned} {\mathcal {V}}_{K}=\left\{ \psi \left( \overline{b_{1}},b_{2},b_{3},b_{4}\right) ,\psi \left( b_{1},\overline{b_{2}},b_{3},b_{4}\right) \right\} . \end{aligned}$$
(27)
Now let us repeat similar derivation for points
\(I,J,N\). According to the
\(M16^{a}\) spectrum we must ensure that
\({\mathbf {v}}_{I}\boxplus {\mathbf {v}}_{D}={\mathbf {v}}_{N}\boxplus {\mathbf {v}}_{D}={\mathbf {v}}_{J}\boxplus {\mathbf {v}}_{D}=1\) and, additionally, that
\({\mathbf {v}}_{J}\boxplus {\mathbf {v}}_{A}\ne 1\). We obtain
$$\begin{aligned} {\mathcal {V}}_{IN}=\left\{ \psi \left( \overline{b_{1}},\overline{b_{2}},\overline{b_{3}},b_{4}\right) ,\psi \left( \overline{b_{1}},\overline{b_{2}},b_{3},\overline{b_{4}}\right) \right\} , \end{aligned}$$
and
$$\begin{aligned} {\mathcal {V}}_{J}=\left\{ \psi \left( b_{1},\overline{b_{2}},b_{3},b_{4}\right) ,\psi \left( \overline{b_{1}},b_{2},b_{3},b_{4}\right) \right\} ={\mathcal {V}}_{K}. \end{aligned}$$
(28)
At this stage we can assign the labels to points
\(M\) and
\(P\). To accomplish this goal we make use of the fact that the spectrum of
\(\alpha \) points contains none of
\(d\),
\(4d\) and
\(10d\) entries. Thus, it is clear that the labels of
\(M\) and
\(P\) cannot differ in one bit from any element of
\(\{{\mathcal {V}}_{IN},{\mathcal {V}}_{LO}\}\). It holds because all
\(\{{\mathcal {V}}_{IN},{\mathcal {V}}_{LO}\}\) entries have to be dispatched to points
\(I,N,L \text{ and } O\). As a result,
$$\begin{aligned} {\mathcal {V}}_{M}={\mathcal {V}}_{P}=\left\{ \psi \left( b_{1},\overline{b_{2}},\overline{b_{3}},\overline{b_{4}}\right) ,\psi \left( \overline{b_{1}},b_{2},\overline{b_{3}},\overline{b_{4}}\right) \right\} \equiv {\mathcal {V}}_{MP}. \end{aligned}$$
(29)
At the end we note that
\(\psi (b_{1},\overline{b_{2}},\overline{b_{3}},\overline{b_{4}})\boxplus \psi (\overline{b_{1}},b_{2},\overline{b_{3}},\overline{b_{4}})=2\),
\({\mathbf {v}}_{A}\boxplus \psi (b_{1},\overline{b_{2}},\overline{b_{3}},\overline{b_{4}})={\mathbf {v}}_{D}\boxplus \psi (\overline{b_{1}},b_{2},\overline{b_{3}},\overline{b_{4}})={\mathbf {v}}_{A}\boxplus \psi (\overline{b_{1}},b_{2},\overline{b_{3}},\overline{b_{4}})={\mathbf {v}}_{D}\boxplus \psi (b_{1},\overline{b_{2}},\overline{b_{3}},\overline{b_{4}})=3\), which proves the theorem, regardless of the way in which the elements of
\({\mathcal {V}}_{MP}\) are finally assigned to points
\(M\) and
\(P\).