We will prove that |𝓖_{i}| ≠ 8 for any *i* ∈ 𝓘 by contradiction. Let us suppose that there exists *i* ∈ 𝓘 such that |𝓖_{i}| = 8. Thus, since

$$\begin{array}{}|{\mathcal{G}}_{i}|=\frac{1}{4}\sum _{\omega \in \mathcal{I}\mathrm{\setminus}\{i,-i\}}|{\mathcal{G}}_{i\omega}|,\end{array}$$

we get

$$\begin{array}{}8=\frac{1}{4}\sum _{\omega \in \mathcal{I}\mathrm{\setminus}\{i,-i\}}|{\mathcal{G}}_{i\omega}|.\end{array}$$

Consequently,

$$\begin{array}{}\sum _{\omega \in \mathcal{I}\mathrm{\setminus}\{i,-i\}}|{\mathcal{G}}_{i\omega}|=32.\end{array}$$(1)

From Proposition 3.5 it follows that |𝓖_{iω}| ≤ 3 for all *ω* ∈ 𝓘 ∖ {*i*, –*i*}. Particular attention will be given to the elements *ω* ∈ 𝓘 ∖ {*i*, –*i*} such that |𝓖_{iω}| = 3 or |𝓖_{iω}| = 2.

Throughout this subsection 𝓙 and 𝓚 will denote the following sets:

$$\begin{array}{}\mathcal{J}=\{j\in \mathcal{I}\mathrm{\setminus}\{i,-i\}:|{\mathcal{G}}_{ij}|=3\}\end{array}$$

and

$$\begin{array}{}\mathcal{K}=\{k\in \mathcal{I}\mathrm{\setminus}\{i,-i\}:|{\mathcal{G}}_{ik}|=2\}.\end{array}$$

We begin by characterizing partially the index distribution of the codewords *w*_{1}, …, *W*_{8} ∈ 𝓖_{i}.

#### Proposition 4.2

*If* |𝓖_{i}| = 8, *i* ∈ 𝓘, *then* 𝓘 ∖ {*i*, –*i*} = 𝓙 ∪ 𝓚, *with* |𝓙| = 8 *and* |𝓚| = 4. *The partial index distribution of the codewords* *w*_{1}, …, *W*_{8} ∈ 𝓖_{i} *satisfies*:

*where* *x*, –*x*, *y*, –*y* ∈ 𝓙 *and k*_{1}, …, *k*_{8} ∈ 𝓚. *Consequently*, *for all* *W* ∈ 𝓖_{i} *there exists a unique element* *k* ∈ 𝓚 *such that* *W* ∈ 𝓖_{ik}.

#### Proof

Let *i* ∈ 𝓘 such that |𝓖_{i}| = 8. In these conditions, (1) is satisfied. By Proposition 3.5, for any *ω* ∈ 𝓘 ∖ {*i*, –*i*} we get |𝓖_{iω}| ≤ 3. As |𝓘 ∖ {*i*, –*i*}| = 12, taking into account (1) we conclude that there are, at least, eight elements *ω* ∈ 𝓘 ∖ {*i*, –*i*} satisfying |𝓖_{iω}| = 3. We have just concluded that |𝓙| ≥ 8.

Let us consider

$$\begin{array}{}\mathcal{L}=\{l\in \mathcal{I}\mathrm{\setminus}\{i,-i\}:|{\mathcal{G}}_{il}|\le 2\}.\end{array}$$

Observing that, 𝓙 ∪ 𝓛 = 𝓘 ∖ {*i*, –*i*}, 𝓙 ∩ 𝓛 = ∅, |𝓘 ∖ {*i*, –*i*}| = 12 and |𝓙| ≥ 8, then |𝓛| ≤ 4. Thus, there are, at most, four distinct elements *j* ∈ 𝓙 such that –*j* ∈ 𝓛. Since |𝓙| ≥ 8, there exist x, y ∈ 𝓙, distinct, such that –*x*, –*y* ∈ 𝓙. Then, let us consider *x*, –*x*, *y*, –*y* ∈ 𝓙.

By definition of 𝓙, |𝓖_{ix}| = |𝓖_{i,–x}| = |𝓖_{iy}| = |𝓖_{i, –y}| = 3. Taking into account Lemma 3.3, the partial index distribution of the codewords *w*_{1}, …, *W*_{8} ∈ 𝓖_{i} must satisfy the conditions presented in the , in which *W*_{1} ∈ 𝓖_{ixy}, *W*_{2} ∈ 𝓖_{i,x,–y} and so on.

Table 2 Partial index distribution of the codewords of 𝓖_{i}.

Looking at *W*_{1} ∈ 𝓖_{ixy}, there are *α*, *β* ∈ 𝓘 ∖ {*i*, –*i*, *x*, –*x*, *y*, –*y*} such that *W*_{1} ∈ 𝓖_{ixyαβ}. Suppose that *α*,*β* ∈ 𝓙, that is, |𝓖_{iα}| = |𝓖_{iβ}| = 3. Talking into account Lemma 3.3, |𝓖_{ixα}| = |𝓖_{iyα}| = |𝓖_{ixβ}| = |𝓖_{iyβ}| = 1. Besides, 𝓖_{ixα} = 𝓖_{iyα} = 𝓖_{ixβ} = 𝓖_{iyβ} = {*W*_{1}}. Since |𝓖_{iα}| = 3, taking into account and Lemma 3.3, 𝓖_{iα} ∖ {*W*_{1}} ⊂ {*W*_{5}, *W*_{6}, *W*_{8}} and 𝓖_{iβ} ∖ {*W*_{1}} ⊂ {*W*_{5}, *W*_{6}, *W*_{8}}. As |𝓖_{iα} ∖ {*W*_{1}}| = |𝓖_{iβ} ∖ {*W*_{1}}| = 2, there exists *W* ∈ {*W*_{5}, *W*_{6}, *W*_{8}} such that *W* ∈ 𝓖_{iαβ}, which contradicts Lemma 3.3 since *W*, *W*_{1} ∈ 𝓖_{iαβ}. Therefore, there exists *l*_{1} ∈ 𝓛 so that *W*_{1} ∈ 𝓖_{ixyl1}. Similarly, there are *l*_{2}, *l*_{4}, *l*_{5} ∈ 𝓛 such that *W*_{2} ∈ 𝓖_{i,x,–y,l2}, *W*_{4} ∈ 𝓖_{i,–x,y,l4} and *W*_{5} ∈ 𝓖_{i,–x,–y,l5}.

Let us consider *W*_{3} ∈ 𝓖_{ix}. Having in view *w*_{1}, *W*_{2} ∈ 𝓖_{ix} and Lemma 3.3, there are *α*, *β*, *γ* ∈ 𝓘∖{*i*,–*i*,*x*,–*x*,*y*,–*y*} so that *W*_{3} ∈ 𝓖_{ixαβγ}. Assume that {*α*, *β*, *γ*} ⊂ 𝓙. Then, |𝓖_{iα}| = |𝓖_{iβ}| = |!𝓖_{iγ}| = 3. Accordingly, considering Lemma 3.3, we get |𝓖_{ixα}| = |𝓖_{ixβ}| = |𝓖_{ixγ}| = 1 and, as a consequence, 𝓖_{ixα} = 𝓖_{ixβ} = 𝓖_{ixγ} = {*W*_{3}}. Taking into account and Lemma 3.3, we obtain: 𝓖_{iα} ∖ {*W*_{3}\\} ⊂ {*W*_{4},…,*W*_{8}}; 𝓖_{iβ} ∖ {*W*_{3}} ⊂ {*W*_{4},…,*W*_{8}}; 𝓖_{iγ} ∖ {*W*_{3}} ⊂ {*W*_{4},…,*W*_{8}}. Since |𝓖_{iα} ∖ {*W*_{3}}| = |𝓖_{iβ} ∖ {*W*_{3}}| = |𝓖_{iγ} ∖ {*W*_{3}}| = 2 and |{*W*_{4},…,*W*_{8}}| = 5, there exists *W* ∈ {*W*_{4},…,*W*_{8}} such that *W* ∈ 𝓖_{iεθ} for *ε*, *θ* ∈ {*α*, *β*, *γ*}, which contradicts Lemma 3.3 since *W*, *W*_{3} ∈ 𝓖_{iεθ}. Thus, there exists *l*_{3} ∈ 𝓛 such that *W*_{3} ∈ 𝓖_{ixl3}. Likewise, there are *l*_{6}, *l*_{7}, *l*_{8} ∈ 𝓛 such that *W*_{6} ∈ 𝓖_{i,–x,l6}, *W*_{7} ∈ 𝓖_{iyl7} and *W*_{8} ∈ 𝓖_{i,–y,l8}.

Therefore, for all *W* ∈ 𝓖_{i} there exists *l* ∈ 𝓛 such that *W* ∈ 𝓖_{il}.

By definition of 𝓛, |𝓖_{il}| ≤ 2 for all *l* ∈ 𝓛. We have concluded before that |𝓛| ≤ 4. Since for any *W* ∈ 𝓖_{i} there exists *l* ∈ 𝓛 such that *W* ∈ 𝓖_{il} and |𝓖_{i}| = 8, we must impose |𝓛| = 4 and |𝓖_{il}| = 2 for any *l* ∈ 𝓛. That is, 𝓚 = {*k* ∈ 𝓘 {*i*, −*i*} : |𝓖_{ik}| = 2} is such that |𝓚| = 4. Consequently, for each *W* ∈ 𝓖_{i} there exists a unique element *k* ∈ 𝓚 such that *W* ∈ 𝓖_{ik}. Furthermore, |𝓙| = 8, 𝓘 {*i*, −*i*} = 𝓙 ∪ 𝓚 and the partial index distribution of the codewords of 𝓖_{i} satisfies the conditions which are given in the statement of this proposition. □

The following result characterizes in more detail the set 𝓚 and, consequently, the set 𝓙.

#### Proposition 4.3

*If* *k* ∈ 𝓚, *then* −*k* ∈ 𝓚.

#### Proof

We are assuming |𝓖_{i}| = 8 for *i* ∈ 𝓘. The partial index distribution of the codewords *W*_{ 1}, …,*W*_{8} ∈ 𝓖_{i} satisfies the conditions enunciated in Proposition 4.2. We recall that, from this proposition it follows that 𝓘 \{*i*, −*i*} = 𝓙 ∪ 𝓚, with |𝓙| = 8 and |𝓚| = 4. Furthermore, {*x*, −*x*, *y*, −*y*} ⊂ 𝓙 and {*k*_{1}, …, *k*_{8}} = 𝓚.

Let us consider 𝓝 = 𝓙 {*x*, −*x*, *y*, −*y*} = {*α*, *β*, *γ*, *δ*}. We note that,

$$\begin{array}{}{\displaystyle \mathcal{I}\mathrm{\setminus}\{i,-i\}=\{{k}_{1},\dots ,{k}_{8}\}\cup \{x,-x,y,-y\}\cup \{\alpha ,\beta ,\gamma ,\delta \}.}\end{array}$$

By Proposition 4.2, for each *W* ∈ 𝓖_{i} there exists a unique element *k* ∈ 𝓚 such that *W* ∈ 𝓖_{ik}. On the other hand, since |𝓖_{ij}| = 3 for all *j* ∈ 𝓙, we have identified all codewords of 𝓖_{ix}, 𝓖_{i, −x}, 𝓖_{iy} and 𝓖_{i}, −*y*}. Thus, to characterize completely the index distribution of all codewords of 𝓖_{i} we must fill in with elements of 𝓝 the empty entries of the table presented in Proposition 4.2.

Consider *W*_{1}, *W*_{2}, *W*_{3} ∈ 𝓖_{ix}, see table in Proposition 4.2. Taking into account Lemma 3.3, the index distribution of the codewords of 𝓖_{ix} must satisfy the conditions in .

Table 3 Partial index distribution of the codewords of 𝓖_{i}.

Let us now consider the codeword *W*_{4} ∈ 𝓖_{i}, *k*_{4, −x, y}. Having in mind Lemma 3.3 we conclude that *W*_{4} ∉ 𝓖_{α}, otherwise we would get *W*_{1},*W*_{4} ∈ 𝓖_{iyα}. Suppose that *W*_{4} ∈ 𝓖_{β}. In these conditions, *W*_{4}, *W*_{2} ∈ 𝓖_{iβ}, with *W*_{4} ∈ 𝓖_{i,k4,−x,y,β} and *W*_{2} ∈ 𝓖_{i,k2,x,−y,β}. Since |𝓖_{iβ}| = 3 (*β* ∈ 𝓙), there exists *W* ∈ 𝓖_{i}\{*W*_{1}, *W*_{2}, *W*_{3}, *W*_{4}} such that *W* ∈ 𝓖_{iβ}. By we verify that *W* ∈ 𝓖_{i,β,−x} ∪ 𝓖_{iβy} ∪ 𝓖_{i,β,−y}. Consequently, taking into account *W*_{2} and *W*_{4}, |𝓖_{iβz}| ≥ 2 for some *z* ∈ {−*x*, *y*, −*y*}, contradicting Lemma 3.3.

Therefore, *W*_{4} ∈ 𝓖_{γ} ∪ 𝓖_{δ}. By a similar reasoning, we are led to the conclusion that *W*_{5} ∈ 𝓖_{γ} ∪ 𝓖_{δ}.

We are assuming *W*_{3} ∈ 𝓖_{ik3xγδ}. As *k*_{3} ∈ 𝓚, by definition of 𝓚 we get |𝓖_{ik3}| = 2 . Thus, there exists *k* ∈ {*k*_{1}, …, *k*_{8}}\{*k*_{3}} such that *k* = *k*_{3}. We note that, *k*_{3} ≠ *k*_{1}, *k*_{2}, otherwise Lemma 3.3 is contradicted. Since *W*_{4}, *W*_{5} ∈ 𝓖_{γ} ∪ 𝓖_{δ}, taking into account Lemma 3.3 we conclude that *k*_{3} ≠ *k*_{4}, *k*_{5}. Therefore, *k* ∈ {*k*_{6}, *k*_{7}, *k*_{8}}. If *k*_{3} = *k*_{7}, then Lemma 3.3 forces *W*_{7} ∈ 𝓖_{ik7y}*αβ*, which is a contradiction, since *W*_{1}, *W*_{7} ∈ 𝓖_{iyα}. Then, *k*_{3} ≠ *k*_{7}. By a similar reasoning we may conclude that *k*_{3} ≠ *k*_{8}. Consequently, *k*_{3} = *k*_{6} and, applying once again Lemma 3.3, we must impose *W*_{6} ∈ 𝓖_{i,k3,−x,α,β}.

Note that |𝓖_{iα}| = |𝓖_{iβ}| = 3. Since *W*_{4}, *W*_{5} ∈ 𝓖_{γ} ∪ 𝓖_{δ}, we must obligate *W*_{7}, *W*_{8} ∈ 𝓖_{α} ∪ 𝓖_{β}. Considering *W*_{1} and *W*_{2}, Lemma 3.3 leads us to conclude that *W*_{7} ∈ 𝓖_{β} and *W*_{8} ∈ 𝓖_{α}.

Accordingly, the partial index distribution of the codewords of 𝓖_{i} satisfies:

Table 4 Partial index distribution of the codewords of 𝓖_{i}.

Note that, as |𝓖_{iγ}| = |𝓖_{iδ}| = 3, the four empty entries of this table must be filled in with *γ* and *δ*. Thus, *W*_{4}, *W*_{5}, *W*_{7}, *W*_{8} ∈ 𝓖_{γ} ∪ 𝓖_{δ}.

Consider the elements of 𝓚. By the analysis of the entries of the previous table, to avoid the contradiction of Lemma 3.3, one should have *k*_{1} = *k*_{5}, *k*_{2} = *k*_{4} and *k*_{7} = *k*_{8}. That is, 𝓚 = {*k*_{1}, *k*_{2}, *k*_{3}, *k*_{7}} and the codewords of 𝓖_{i} are characterize as it is presented in .

Table 5 Partial index distribution of the codewords of 𝓖_{i}.

We intend to show that if *k* ∈ 𝓚, then −*k* ∈ 𝓚. Let us focus our attention on *k*_{3} ∈ 𝓚. We have concluded before that *W*_{3}, *W*_{6} ∈ 𝓖_{ik3}, with *W*_{3} ∈ 𝓖_{ik3xγδ} and *W*_{6} ∈ 𝓖_{i,k3,−x,α,β}. In these conditions, −*k*_{3} ∈ 𝓘 \({*i*,−*i*,*x*, −*x*,*y*,−*y*} ∪ 𝓝). That is, −*k*_{3} ∈ 𝓘\({*i*,−*i*} ∪ 𝓙). Since 𝓘 = {*i*, −*i*} ∪ 𝓙 ∪ 𝓚, then −*k*_{3} ∈ 𝓚.

Looking at the codewords *W*_{7}, *W*_{8} ∈ 𝓖_{ik7}, we get *W*_{7} ∈ 𝓖_{γ} and *W*_{8} ∈ 𝓖_{δ}, or, *W*_{7} ∈ 𝓖_{δ} and *W*_{8} ∈ 𝓖_{γ}. In both cases −*k*_{7} ∈ 𝓘\({*i*, −*i*} ∪ 𝓙), accordingly −*k*_{7} ∈ 𝓚.

Now, 𝓚 = {*k*_{1}, *k*_{2}, *k*_{3}, *k*_{7}} and −*k*_{3},-k_{7} ∈ 𝓚. Either *k*_{3} ≠−*k*_{7} or *k*_{3} = −*k*_{7}.

If *k*_{3} ≠ −*k*_{7}, then −*k* ∈ 𝓚 for all *k* ∈ 𝓚.

If *k*_{3} = −*k*_{7} and *k*_{1} = −*k*_{2}, then −*k* ∈ 𝓚 for all *k* ∈ 𝓚.

Assume that *k*_{3} = −*k*_{7} and *k*_{1} ≠ −*k*_{2}. By this assumption it follows that −*k*_{1}, −*k*_{2} ∈ 𝓝 = {*α*, *β*, *γ*, *δ*}. Thus, there are *ε*_{1}, *ε*_{2} ∈ 𝓝 so that −*k*_{1} = *ε*_{1}, −*k*_{2} = *ε*_{2} and the remaining elements of 𝓝, *ε*_{3} and *ε*_{4}, satisfy *ε*_{3} = −*ε*_{4}. As *W*_{1} ∈ 𝓖_{ik1xyα}, then −*k*_{1} ∈ {*β*, *γ*, *δ*}. On the other hand, since *W*_{2} ∈ 𝓖_{i,k2,x,−y,β}, then −*k*_{2} ∈ {*α*,*γ*,*δ*}. We note that, as *k*_{1} ≠ *k*_{2}, then −*k*_{1} ≠ −*k*_{2}.

If −*k*_{1} = *β* and −*k*_{2} = *α*, then *γ* = −*δ*, which is a contradiction since *W*_{3} ∈ 𝓖_{ik3xγδ}.

If −*k*_{1} = *β* and −*k*_{2} = *γ*, then *α* = −*δ*. Analyzing and taking into account that *W*_{4} ∈ 𝓖_{γ} ∪ 𝓖_{δ}, we conclude that *W*_{4} ∈ 𝓖_{i,k2,−x,y,δ}. Consequently, having in mind Lemma 3.3, *W*_{5} ∈ 𝓖_{i,k1,−x,−y,γ}, *W*_{7} ∈ 𝓖_{ik7yβγ} and *W*_{8} ∈ 𝓖_{i,k7,−y,α,δ}, which is not possible since we are supposing *α* = −*δ*.

If −*k*_{1} = *β* and −*k*_{2} = *δ*, then *α* = −*γ*. Consequently, *W*_{8} ∈ 𝓖_{i,k7,−y,α,δ}, *W*_{7} ∈ 𝓖_{ik7yβγ} and *W*_{4} ∈ 𝓖_{i,k2,−x,y,δ}. We get a contradiction since, by hypothesis, −*k*_{2} = *δ*.

Combining all possibilities for −*k*_{1} ∈ {*β*, *γ*, *δ*} and −*k*_{2} ∈ {*α*, *γ*, *δ*}, by a similar reasoning we get always a contradiction. Therefore, −*k* ∈ 𝓚 for all *k* ∈ 𝓚. □

From Proposition 4.2 we get 𝓘\{*i*, −*i*} = 𝓙 ∪ 𝓚. We have just seen that, if *k* ∈ 𝓚 then −*k* ∈ 𝓚. So, if *j* ∈ 𝓙 then −*j* ∈ 𝓙.

Until this moment we have focused our attention on the characterization of the codewords of 𝓖_{i}. The two following propositions arise from the analysis of other type of codewords, in particular, codewords of 𝓓 ∪ 𝓔 ∪ 𝓕.

#### Proposition 4.4

*If* |𝓖_{i}| = 8, *i* ∈ 𝓘, *then* |𝓕_{i}| = 0.

#### Proof

Let |𝓖_{i}| = 8 for *i* ∈ 𝓘. Suppose, by contradiction, that |𝓕_{i}| > 0. Let *U* ∈ 𝓕_{i}. Since the codewords of 𝓕 are of type [±2, ±1^{3}], there exist $u_{1}, u_{2}, u_{3} ∈ 𝓘 {*i*, −*i*}, with |*u*_{1}|, |*u*_{2}| and |*u*_{3}| pairwise distinct, such that *U* ∈ 𝓕_{iu1u2u3}.

By Proposition 4.2, 𝓘\{*i*, −*i*} = 𝓙 ∪ 𝓚, therefore *u*_{1}, *u*_{2}, *u*_{3} ∈ 𝓙 ∪ 𝓚. Recall that |𝓖_{ij}| = 3 for any *j* ∈ 𝓙. Then, by Proposition 3.5 one has |𝓕_{ij}| = 0 for all *j* ∈ 𝓙. Consequently, *u*_{1}, *u*_{2}, *u*_{3} ∈ 𝓚. From Proposition 4.2 it follows that |𝓚| = 4 and, taking into account Proposition 4.3, −*k* ∈ 𝓚 for all *k* ∈ 𝓚. Thus, is not possible to have *u*_{1}, *u*_{2}, *u*_{3} ∈ 𝓚 satisfying |*u*_{1}|, |*u*_{2}| and |*u*_{3}| pairwise distinct, contradicting our assumption.□

#### Proposition 4.5

*For all* *j* ∈ 𝓙, |𝓓_{ij} ∪ 𝓔_{ij}| = 1. *For all* *k* ∈ 𝓚, |𝓓_{ik} ∪ 𝓔_{ik}| = 4. *Furthermore*, *if* *k* ∈ 𝓚, *the codewords* *U*_{1}, *U*_{2}, *U*_{3}, *U*_{4} ∈ 𝓓_{ik} ∪ 𝓔_{ik} *are such that* *U*_{1} ∈ 𝓓_{iku1} ∪ 𝓔_{iku1}, *U*_{2} ∈ 𝓓_{iku2} ∪ 𝓔_{iku2}, *U*_{3} ∈ 𝓓_{iku3} ∪ 𝓔_{iku3} and *U*_{4} ∈ 𝓓_{iku4} ∪ 𝓔_{iku4}, *with u*_{1}, *u*_{2} ∈ 𝓙, *u*_{1} ≠ *u*_{2}, *and* *u*_{3}, *u*_{4} ∈ 𝓚 {*k*, −*k*}, *with* *u*_{3} = −*u*_{4}.

#### Proof

From Proposition 3.5 we get

$$\begin{array}{}{\displaystyle |{\mathcal{D}}_{i\omega}\cup {\mathcal{E}}_{i\omega}|+2|{\mathcal{F}}_{i\omega}|+3|{\mathcal{G}}_{i\omega}|=10}\end{array}$$(2)

for all *ω* ∈ 𝓘\{*i*, −*i*}. By Proposition 4.4 we know that |𝓕_{i}| = 0 and, consequently, |𝓕_{iω}| = 0 for all *ω* ∈ 𝓘\{*i*, −*i*}. As |𝓖_{ij}| = 3 for any *j* ∈ 𝓙, from (2) we obtain |𝓓_{ij} ∪ 𝓔_{ij}| = 1 for all *j* ∈ 𝓙. Considering again (2), we conclude that |𝓓_{ik} ∪ 𝓔_{ik}| = 4 for each *k* ∈ 𝓚, since |𝓖_{ik}| = 2 for all *k* ∈ 𝓚.

Let *k* ∈ 𝓚. Then, there exist *V*_{1}, *V*_{2} ∈ 𝓖_{ik} and *U*_{1}, …, *U*_{4} ∈ 𝓓_{ik} ∪ 𝓔_{ik}. We note that, the codewords of 𝓓 are of type [±3,±1^{2}] and the codewords of 𝓔 are of type [±2^{2}, ±1]. Thus, there are *v*_{1}, …, *v*_{6}, *u*_{1}, …, *u*_{4} in 𝓘 {*i*, −*i*, *k*, −*k*} such that:

Table 6 Index distribution of the codewords of 𝓖_{ik} ∪ 𝓓_{ik} ∪ 𝓔_{ik}.

It should be pointed out that, by Lemma 3.3, *v*_{1}, …, *v*_{6}, *u*_{1}, …, *u*_{4} must be pairwise distinct. Therefore, {*v*_{1}, …, *v*_{6}, *u*_{1}, …, *u*_{4}} = 𝓘 {*i*, −*i*, *k*, −*k*}. By Proposition 4.2, 𝓘 {*i*, −*i*} = 𝓙 ∪ 𝓚, with |𝓙| = 8 and |𝓚| = 4. Furthermore, from Proposition 4.2, −*k* ∈ 𝓚. Then, {*v*_{1}, …, *v*_{6}, *u*_{1}, …, *u*_{4}} = 𝓙 ∪ 𝓚 {*k*, −*k*}. Since *V*_{1},*V*_{2} ∈ 𝓖_{ik}, with *k* ∈ 𝓚, taking into account Proposition 4.2 we must impose {*v*_{1}, …, *v*_{6}} ⊂ 𝓙. Consequently, without loss of generality, *u*_{1}, *u*_{2} ∈ 𝓙 and *u*_{3}, *u*_{4} ∈ 𝓚 {*k*, −*k*}. Considering Proposition 4.2 we conclude that *u*_{3} = −*u*_{4}. □

We are now able to establish the main result of this paper.

#### Theorem 4.6

*For any* *i* ∈ 𝓘, |𝓖_{i}| ≠ 8.

#### Proof

By contradiction, consider *i* ∈ 𝓘 such that |𝓖_{i}| = 8.

From Proposition 4.2 we have |𝓚| = 4, so let *k* be an element of 𝓚. By Proposition 4.5, there exist *U*_{1}, …, *U*_{4} ∈ 𝓓_{ik} ∪ 𝓔_{ik} whose index distribution satisfies the conditions presented in , where *u*, −*u* ∈ 𝓚\{*k*, −*k*} and *j*_{1}, *j*_{2} ∈ 𝓙, with *j*_{1} ≠ *j*_{2}. We note that, in these conditions, 𝓚 = {*k*, −*k*, *u*, −*u*}.

Table 7 Index distribution of the codewords of 𝓓_{ik} ∪ 𝓔_{ik}.

Let us denote by 𝓗 the set of words of type [±2, ±1]. Consider the words *P*_{1}, *P*_{2} ∈ 𝓗 such that *P*_{1} ∈
$\begin{array}{}{\displaystyle {\mathcal{H}}_{i}^{(2)}\cap {\mathcal{H}}_{{j}_{1}}^{(1)}}\end{array}$

and *P*_{2} ∈
$\begin{array}{}{\displaystyle {\mathcal{H}}_{i}^{(2)}\cap {\mathcal{H}}_{{j}_{2}}^{(1)}}\end{array}$
. The index distribution of the codewords of 𝓓_{ik} ∪ 𝓔_{ik} and the index value distribution of the words *P*_{1} and *P*_{2} are represented in the following table:

Table 8 Index distribution of *U*_{1}, …, *U*_{4} ∈ 𝓓_{ik} ∪ 𝓔_{ik} and index value distribution of *P*_{1}, *P*_{2} ∈ 𝓗_{i}.

By definition of perfect 2-error correcting Lee code, for each *P* ∈ {*P*_{1}, *P*_{2}} there exists a unique codeword *V* ∈ 𝓣 such that *μ*_{L}(*P*, *V*) ≤ 2. Taking into account the type of words of 𝓗 as well as the fact of |𝓕_{i}| = 0 (see Proposition 4.4), each word *P*_{q}
$\begin{array}{}{\displaystyle {\mathcal{H}}_{i}^{(2)}\cap {\mathcal{H}}_{{j}_{q}},}\end{array}$
with *j*_{q} ∈ 𝓘\{*i*, −*i*}, is covered by a unique codeword

$$\begin{array}{}{\displaystyle {V}_{q}\in ({\mathcal{B}}_{i}^{(4)}\cap {\mathcal{B}}_{{j}_{q}}^{(1)})\cup {\mathcal{C}}_{i{j}_{q}}\cup ({\mathcal{D}}_{i}^{(3)}\cap {\mathcal{D}}_{{j}_{q}}^{(1)})\cup ({\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{{j}_{q}}).}\end{array}$$(3)

Thus, we may consider *U*_{3} and *U*_{4} as possible codewords to cover *P*_{1} and *P*_{2}, respectively.

Suppose that *P*_{1} is covered by *U*_{3} and *P*_{2} is covered by *U*_{4}. Then, we must impose

$$\begin{array}{}{\displaystyle {U}_{3}\in ({\mathcal{D}}_{i}^{(3)}\cap {\mathcal{D}}_{k}^{(1)}\cap {\mathcal{D}}_{{j}_{1}}^{(1)})\cup ({\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{k}\cap {\mathcal{E}}_{{j}_{1}})}\end{array}$$

and

$$\begin{array}{}{\displaystyle {U}_{4}\in ({\mathcal{D}}_{i}^{(3)}\cap {\mathcal{D}}_{k}^{(1)}\cap {\mathcal{D}}_{{j}_{2}}^{(1)})\cup ({\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{k}\cap {\mathcal{E}}_{{j}_{2}}),}\end{array}$$

which contradicts Lemma 3.2, since *U*_{3},*U*_{4} ∈
$\begin{array}{}({\mathcal{D}}_{i}^{(3)}\cap {\mathcal{D}}_{k}^{(1)})\cup ({\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{k}).\end{array}$

Therefore, either *P*_{1} is not covered by *U*_{3} or *P*_{2} is not covered by *U*_{4}.

Without loss of generality, let us assume that *P*_{1} is not covered by *U*_{3}. Note that, *U*_{3} ∈ 𝓓_{ikj1} ∪ 𝓔_{ikj1}. As *j*_{1} ∈ 𝓙, by Proposition 4.5 we get |𝓓_{ij1} ∪ 𝓔_{ij1}| = 1. Consequently, 𝓓_{ij1} ∪ 𝓔_{ij1} = {*U*_{3}. Since we are assuming that *U*_{3} does not cover *P*_{1}, considering (3), *P*_{1} is covered by a codeword *V*_{1} satisfying *V*_{1} ∈
$\begin{array}{}({\mathcal{B}}_{i}^{(4)}\cap {\mathcal{B}}_{{j}_{1}}^{(1)})\cup {\mathcal{C}}_{i{j}_{1}}.\end{array}$

Next, we will analyze, separately, the hypotheses:

*V*_{1} ∈
$\begin{array}{}{\mathcal{B}}_{i}^{(4)}\cap {\mathcal{B}}_{{j}_{1}}^{(1)}\end{array}$;

*V*_{1} ∈ 𝓒_{ij1}.

Assume that *P*_{1} is covered by *V*_{1} ∈
$\begin{array}{}{\mathcal{B}}_{i}^{(4)}\cap {\mathcal{B}}_{{j}_{1}}^{(1)}\end{array}$.

Assuming that *P*_{1} is covered by *V*_{1} ∈
$\begin{array}{}{\mathcal{B}}_{i}^{(4)}\cap {\mathcal{B}}_{{j}_{1}}^{(1)}\end{array}$, by Lemma 3.1 we conclude
$\begin{array}{}|{\mathcal{B}}_{i}^{(4)}\mathrm{\setminus}\{{V}_{1}\}\cup {\mathcal{C}}_{i}^{(3)}\cup {\mathcal{D}}_{i}^{(3)}|=0.\end{array}$ Consequently, if *U* ∈ {*U*_{1},…,*U*_{4} is such that *U* ∈ 𝓓, then
$\begin{array}{}U\in {\mathcal{D}}_{i}^{(1)}.\end{array}$ Furthermore, *P*_{2} must be covered by

$$\begin{array}{}{V}_{2}\in ({\mathcal{C}}_{i}^{(2)}\cap {\mathcal{C}}_{{j}_{2}}^{(3)})\cup ({\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{{j}_{2}}).\end{array}$$

If *V*_{2} ∈
$\begin{array}{}{\mathcal{E}}_{i}^{(2)}\end{array}$ ∩ 𝓔_{j2}, since *j*_{2} ∈ 𝓙 we conclude, by Proposition 4.5, that *V*_{2} = *U*_{4}. Having in mind *U*_{1}, *U*_{2} and *U*_{3}, see , if *U* ∈ {*U*_{1}, *U*_{2}, *U*_{3}} is such that *U* ∈ 𝓔, then *U* ∈
$\begin{array}{}{\mathcal{E}}_{i}^{(1)}\end{array}$, otherwise, *U*, *U*_{4} ∈
$\begin{array}{}{\mathcal{E}}_{i}^{(2)}\end{array}$ ∩ 𝓔_{k}, contradicting Lemma 3.2. Therefore, since we have concluded before that {*U*_{1}, *U*_{2}, *U*_{3}} ∩
$\begin{array}{}{\mathcal{D}}_{i}^{(3)}\end{array}$ = ∅, we get *U*_{1}, *U*_{2}, *U*_{3} ∈
$\begin{array}{}{\mathcal{D}}_{i}^{(1)}\cup {\mathcal{E}}_{i}^{(1)}\end{array}$. Taking into account the index distribution of *U*_{1} and *U*_{2}, we must have
$\begin{array}{}{U}_{1}\in {\mathcal{D}}_{u}^{(3)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{or}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{U}_{2}\in {\mathcal{D}}_{-u}^{(3)},\end{array}$ otherwise we get *U*_{1}, *U*_{2} ∈
$\begin{array}{}({\mathcal{D}}_{i}^{(1)}\cap {\mathcal{D}}_{k}^{(3)})\cup ({\mathcal{E}}_{i}^{(1)}\cap {\mathcal{E}}_{k}^{(2)}),\end{array}$ contradicting, once again, Lemma 3.2.

If
$\begin{array}{}{V}_{2}\in {\mathcal{C}}_{i}^{(2)}\cap {\mathcal{C}}_{{j}_{2}}^{(3)},\end{array}$ to avoid the contradiction of Lemma 3.2 we must impose
$\begin{array}{}{U}_{4}\in {\mathcal{D}}_{k}^{(3)}.\end{array}$ Consequently, considering again Lemma 3.2, *U*_{1}, *U*_{2}, *U*_{3} ∈
$\begin{array}{}{\mathcal{D}}_{k}^{(1)}\cup {\mathcal{E}}_{k}^{(1)}.\end{array}$ We recall that, we have seen before that {*U*_{1}, *U*_{2}, *U*_{3}} ∩
$\begin{array}{}{\mathcal{D}}_{i}^{(3)}=\varnothing .\end{array}$ Thus, in these conditions,
$\begin{array}{}{U}_{1}\in {\mathcal{D}}_{u}^{(3)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{or}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{U}_{2}\in {\mathcal{D}}_{-u}^{(3)},\end{array}$ otherwise, *U*_{1}, *U*_{2} ∈
$\begin{array}{}{\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{k}^{(1)},\end{array}$ contradicting again Lemma 3.2.

Therefore, in both cases, supposing *V*_{2} ∈
$\begin{array}{}{\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{{j}_{2}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{or}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{V}_{2}\in {\mathcal{C}}_{i}^{(2)}\cap {\mathcal{C}}_{{j}_{2}}^{(3)},\end{array}$ we conclude that
$\begin{array}{}{U}_{1}\in {\mathcal{D}}_{u}^{(3)}\end{array}$ or
$\begin{array}{}{U}_{2}\in {\mathcal{D}}_{-u}^{(3)}.\end{array}$

Suppose, without loss of generality, that
$\begin{array}{}{U}_{1}\in {\mathcal{D}}_{u}^{(3)}\end{array}$. As *u* ∈ 𝓚, by Proposition 4.5 there are *U*_{5}, *U*_{6} ∈ 𝓓_{iu}∪𝓔_{iu} satisfying *U*_{5} ∈ 𝓓_{iuj3} ∪ 𝓔_{iuj3} and *U*_{6} ∈ 𝓓_{iuj4} ∪ 𝓔_{iuj4}, with *j*_{3}, *j*_{4} ∈ 𝓙 distinct. Note that, *j*_{1},…,*j*_{4} ∈ 𝓙 are pairwise distinct, since by Proposition 4.5 we have |𝓓_{ij} ∪ 𝓔_{ij}| = 1 for all *j* ∈ 𝓙.

Let us consider
$\begin{array}{}{P}_{3}\in {\mathcal{H}}_{i}^{(2)}\cap {\mathcal{H}}_{{j}_{3}}^{(1)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{P}_{4}\in {\mathcal{H}}_{i}^{(2)}\cap {\mathcal{H}}_{{j}_{4}}^{(1)}.\end{array}$ summarizes the conditions that the index distribution, and, in some cases, the index value distribution, of the codewords and words described until now, must satisfy.

Table 9 Index conditions on 𝓑_{i} ∪ 𝓓_{i} ∪ 𝓔_{i} and on 4 words of type [±2, ±1].

Taking into account the words *P*_{3} and *P*_{4} we may conclude, as we have concluded before for *P*_{1} and *P*_{2}, that either *P*_{3} is not covered by *U*_{5} or *P*_{4} is not covered by *U*_{6}. In fact, if *U*_{5} covers *P*_{3} and *U*_{6} covers *P*_{4}, then *U*_{5}, *U*_{6} ∈
$\begin{array}{}({\mathcal{D}}_{i}^{(3)}\cap {\mathcal{D}}_{u}^{(1)})\cup ({\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{u}),\end{array}$ contradicting Lemma 3.2. Let us assume, without loss of generality, that *P*_{3} is not covered by *U*_{5}. By Proposition 4.5 it follows that |𝓓_{ij3} ∪ 𝓔_{ij3}| = 1. Consequently, 𝓓_{ij3} ∪ 𝓔_{ij3} = {*U*_{5}}. As a consequence of the assumption *V*_{1} ∈
$\begin{array}{}{\mathcal{B}}_{i}^{(4)}\cap {\mathcal{B}}_{{j}_{1}}^{(1)}\end{array}$ we get
$\begin{array}{}|{\mathcal{B}}_{i}^{(4)}\mathrm{\setminus}\{{V}_{1}\}\cup {\mathcal{C}}_{i}^{(3)}\cup {\mathcal{D}}_{i}^{(3)}|=0.\end{array}$ Thus, under these conditions and taking into account (3), *P*_{3} must be covered by a codeword *V*_{3} satisfying *V*_{3} ∈
$\begin{array}{}{\mathcal{C}}_{i}^{(2)}\cap {\mathcal{C}}_{{j}_{3}}^{(3)}.\end{array}$ Consequently,
$\begin{array}{}{U}_{5}\in {\mathcal{D}}_{u}^{(3)},\end{array}$ otherwise, *U*_{5} ∈
$\begin{array}{}({\mathcal{D}}_{i}^{(1)}\cap {\mathcal{D}}_{{j}_{3}}^{(3)})\cup ({\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{{j}_{3}})\cup ({\mathcal{E}}_{i}\cap {\mathcal{E}}_{{j}_{3}}^{(2)})\end{array}$ and contradicts with the codeword *V*_{3} Lemma 3.2. However, *U*_{1},
$\begin{array}{}{U}_{5}\in {\mathcal{D}}_{u}^{(3)},\end{array}$ contradicting Lemma 3.1.

Accordingly, *P*_{1} can not be covered by the codeword *V*_{1} ∈
$\begin{array}{}{\mathcal{B}}_{i}^{(4)}\cap {\mathcal{B}}_{{j}_{1}}^{(1)}.\end{array}$

**Assume that** *P*_{1} **is covered by** *V*_{1} ∈ 𝓒_{ij1}.

Since *V*_{1} ∈ 𝓒, then *V*_{1} is a codeword of type [±3, ±2]. According with what is being supposed, *V*_{1} ∈
$\begin{array}{}{\mathcal{C}}_{i}^{(3)}\cap {\mathcal{C}}_{{j}_{1}}^{(2)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{or}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{V}_{1}\in {\mathcal{C}}_{i}^{(2)}\cap {\mathcal{C}}_{{j}_{1}}^{(3)}.\end{array}$ Consider *U*_{3} ∈ 𝓓_{ikj1} ∪ 𝓔_{ikj1}. In order to have Lemma 3.2 fulfilled we must force *U*_{3} ∈
$\begin{array}{}{\mathcal{D}}_{i}^{(1)}\cap {\mathcal{D}}_{k}^{(3)}\cap {\mathcal{D}}_{{j}_{1}}^{(1)}.\end{array}$ Schematically, we get .

Table 10 Index distribution on 𝓒_{i} ∪ 𝓓_{i} ∪ 𝓔_{i} and on 2 words of type [±2, ±1].

Taking into account *U*_{3}, by Lemma 3.2 we must have *U*_{1}, *U*_{2}, *U*_{4} ∈
$\begin{array}{}{\mathcal{D}}_{k}^{(1)}\cup {\mathcal{E}}_{k}^{(1)}.\end{array}$ Besides,
$\begin{array}{}{U}_{1}\in {\mathcal{D}}_{u}^{(3)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{or}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{U}_{2}\in {\mathcal{D}}_{-u}^{(3)},\end{array}$ otherwise, *U*_{1}, *U*_{2} ∈
$\begin{array}{}({\mathcal{D}}_{i}^{(3)}\cap {\mathcal{D}}_{k}^{(1)})\cup ({\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{k}^{(1)}),\end{array}$ contradicting Lemma 3.2.

Let us assume, without loss of generality, that
$\begin{array}{}{U}_{1}\in {\mathcal{D}}_{u}^{(3)},\end{array}$

Proceeding as in the previous case, we will consider *U*_{5} ∈ 𝓓_{iuj3} ∪ 𝓔_{iuj3} and *U*_{6} ∈ 𝓓_{iuj4} ∪ 𝓔_{iuj4}, with *j*_{3}, *j*_{4} ∈ 𝓙 and distinct. We will consider also *P*_{3} ∈
$\begin{array}{}{\mathcal{H}}_{i}^{(2)}\cap {\mathcal{H}}_{{j}_{3}}^{(1)}\end{array}$ and *P*_{4} ∈
$\begin{array}{}{\mathcal{H}}_{i}^{(2)}\cap {\mathcal{H}}_{{j}_{4}}^{(1)}\end{array}$. Gathering the information obtained so far, one has the index distribution presented in .

Table 11 Index distribution on 𝓒_{i} ∪ 𝓓_{i} ∪ 𝓔_{i} and on 4 words of type [±2, ±1].

As seen in the previous case, either *U*_{5} does not cover *P*_{3} or *U*_{6} does not cover *P*_{4}. Assume, without loss of generality, that *P*_{3} is not covered by *U*_{5}. By Proposition 4.5 we get 𝓓_{ij3} ∪ 𝓔_{ij3} = {*U*_{5}}. Therefore, considering (3), *P*_{3} must be covered by a codeword *V*_{3} ∈
$\begin{array}{}({\mathcal{B}}_{i}^{(4)}\cap {\mathcal{B}}_{{j}_{3}}^{(1)})\end{array}$ ∪ 𝓒_{ij3}. If *V*_{3} ∈ 𝓒_{ij3}, then, by Lemma 3.2, we must impose *U*_{5} ∈
$\begin{array}{}{\mathcal{D}}_{u}^{(3)}\end{array}$ and, consequently, |
$\begin{array}{}{\mathcal{D}}_{u}^{(3)}\end{array}$| ≥ 2, contradicting Lemma 3.1. Accordingly, *V*_{3} ∈
$\begin{array}{}({\mathcal{B}}_{i}^{(4)}\cap {\mathcal{B}}_{{j}_{3}}^{(1)})\end{array}$.

Taking into account Lemma 3.1,
$\begin{array}{}|{\mathcal{B}}_{i}^{(4)}\mathrm{\setminus}\{{V}_{3}\}\cup {\mathcal{C}}_{i}^{(3)}\cup {\mathcal{D}}_{i}^{(3)}|=0.\end{array}$ Thus, by (3) we may conclude that *P*_{4} must be covered by a codeword

$$\begin{array}{}{V}_{4}\in ({\mathcal{C}}_{i}^{(2)}\cap {\mathcal{C}}_{{j}_{4}}^{(3)})\cup ({\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{{j}_{4}}).\end{array}$$

Note that, if
$\begin{array}{}{V}_{4}\in {\mathcal{C}}_{i}^{(2)}\cap {\mathcal{C}}_{{j}_{4}}^{(3)},\end{array}$ then, by Lemma 3.2, *U*_{6} ∈
$\begin{array}{}{\mathcal{D}}_{u}^{(3)}\end{array}$ implying |
$\begin{array}{}{\mathcal{D}}_{u}^{(3)}\end{array}$| ≥ 2 and contradicting Lemma 3.1. Thus, *V*_{4} ∈
$\begin{array}{}{\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{{j}_{4}}.\end{array}$ By Proposition 4.5, |𝓓_{ij4} ∪ 𝓔_{ij4}| = 1 leading to 𝓓_{ij4} ∪ 𝓔_{ij4} = {*U*_{6} and, consequently, *V*_{4} = *U*_{6}. Since *U*_{1} ∈
$\begin{array}{}{\mathcal{D}}_{i}^{(1)}\cap {\mathcal{D}}_{k}^{(1)}\cap {\mathcal{D}}_{u}^{(3)},\end{array}$ taking into account Lemma 3.2, we must force *U*_{6} ∈
$\begin{array}{}{\mathcal{E}}_{i}^{(2)}\cap {\mathcal{E}}_{u}^{(1)}\cap {\mathcal{E}}_{{j}_{4}}^{(2)}.\end{array}$ The index distribution, and, in some cases the index value distribution, of the codewords and words which we are dealing with are presented in .

Table 12 Index distribution on 𝓑_{i} ∪ 𝓒_{i} ∪ 𝓓_{i} ∪ 𝓔_{i} and on 4 words of type [±2, ±1].

Let us now focus our attention on –*u* ∈ 𝓚. By Proposition 4.5, there are codewords *U*_{7}, *U*_{8} ∈ 𝓓_{i,–u} ∪ 𝓔_{i,–u}, so that, *U*_{7} ∈ 𝓓_{i,–u,j5} ∪ 𝓔_{i,–u,j5} and *U*_{8} ∈ 𝓓_{i,–u,j6} ∪ 𝓔_{i,–u,j6}, with *j*_{5}, *j*_{6} ∈ 𝓙 distinct. Note that, by Proposition 4.5, |𝓓_{ij} ∪ 𝓔_{ij}| = 1 for all *j* ∈ 𝓙, and so *j*_{1},…,*j*_{6} are pairwise distinct. Taking into account the existence of the words *P*_{5} ∈
$\begin{array}{}{\mathcal{H}}_{i}^{(2)}\cap {\mathcal{H}}_{{j}_{5}}^{(1)}\end{array}$ and *P*_{6} ∈
$\begin{array}{}{\mathcal{H}}_{i}^{(2)}\cap {\mathcal{H}}_{{j}_{6}}^{(1)}\end{array}$, we obtain the index distribution presented schematically in .

Table 13 Index distribution on 𝓑_{i} ∪ 𝓒_{i} ∪ 𝓓_{i} ∪ 𝓔_{i} and on 6 words of type [±2, ±1].

By a similar reasoning to the one done with the words *P*_{1}, *P*_{2}, *P*_{3}, *P*_{4} ∈
$\begin{array}{}{\mathcal{H}}_{i}^{(2)},\end{array}$ we conclude that either *P*_{5} is not covered by *U*_{7} or *P*_{6} is not covered by *U*_{8}. Let us assume, without loss of generality, that *U*_{7} does not cover *P*_{5}. Then, considering (3) we are lead to conclude that *P*_{5} must be covered by a codeword

$$\begin{array}{}{P}_{5}\in ({\mathcal{B}}_{i}^{(4)}\cap {\mathcal{B}}_{{j}_{5}}^{(1)})\cup ({\mathcal{C}}_{i{j}_{5}}).\end{array}$$

As *V*_{3} ∈
$\begin{array}{}{\mathcal{B}}_{i}^{(4)}\cap {\mathcal{B}}_{{j}_{3}}^{(1)},\end{array}$ by Lemma 3.1, *P*_{5} ∈
$\begin{array}{}{\mathcal{C}}_{i}^{(2)}\cap {\mathcal{C}}_{{j}_{5}}^{(3)}.\end{array}$ Consequently, taking into account Lemma 3.2, we must force
$\begin{array}{}{U}_{7}\in {\mathcal{D}}_{-u}^{(3)}.\end{array}$

Focus our attention on the codeword *U*_{2} ∈ 𝓓_{i,k,–u} ∪ 𝓔_{i,k,–u}. Having in mind the index value distribution of the codewords *V*_{3}, *U*_{3} and *U*_{7} and considering Lemma 3.1, we conclude that *U*_{2} ∈ 𝓔_{i}. Consequently, either *U*_{2} ∈ 𝓔_{i} ∩
$\begin{array}{}{\mathcal{E}}_{k}^{(2)}\end{array}$ or *U*_{2} ∈ 𝓔_{i} ∩
$\begin{array}{}{\mathcal{E}}_{-u}^{(2)}.\end{array}$ If *U*_{2} ∈ 𝓔_{i} ∩
$\begin{array}{}{\mathcal{E}}_{k}^{(2)}\end{array}$, then the index value distribution of *U*_{2} and *U*_{3} contradicts Lemma 3.2. If *U*_{2} ∈ 𝓔_{i} ∩
$\begin{array}{}{\mathcal{E}}_{-u}^{(2)},\end{array}$ the index value distribution of *U*_{2} and *U*_{7} contradicts also Lemma 3.2.

In both hypotheses, *P*_{1} covered by *V*_{1} ∈
$\begin{array}{}{\mathcal{B}}_{i}^{(4)}\cap {\mathcal{B}}_{{j}_{1}}^{(1)}\end{array}$ or *P*_{1} covered by *V*_{1} ∈ 𝓒_{ij1}, we get a contradiction. □

We have proved in [16] that for each *i* ∈ 𝓘, 3 ≤ |𝓖_{i}| ≤ 8. From last theorem it follows immediately:

#### Corollary 4.7

*For any* *i* ∈ 𝓘, 3 ≤ |𝓖_{i}| ≤ 7.

Since *g* = |𝓖| =
$\begin{array}{}\frac{1}{5}\sum _{i\in \mathcal{I}}|{\mathcal{G}}_{i}|,\end{array}$ the required solutions for the system of
equations presented in Proposition 3.4 must satisfy 9 ≤ *g* ≤ 19. As we have said before, our strategy to prove the non-existence of PL(7, 2) codes consists in getting a minimum range for the variation of |𝓖_{i}|, with *i* ∈ 𝓘, and consequently to reduce the number of solutions for the referred system of equations.

We have already started working on the analysis of other values for |𝓖_{i}|, with *i* ∈ 𝓘, which brings increased difficulties, imposing new strategies and techniques. It seems that our intuition on the new strategy to be applied (from now on) for the proving of the non-existence of PL(7, 2) codes will be successful.

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