Restriction conditions on PL(7, 2) codes (3 ≤ |𝓖_i| ≤ 7)

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 codewordsw₁, …, W₈ ∈ 𝓖_isatisfies:

wherex, –x, y, –y ∈ 𝓙 and k₁, …, k₈ ∈ 𝓚. Consequently, for allW ∈ 𝓖_ithere exists a unique elementk ∈ 𝓚 such thatW ∈ 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

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 w₁, …, W₈ ∈ 𝓖_i must satisfy the conditions presented in the Table 2, in which W₁ ∈ 𝓖_ixy, W₂ ∈ 𝓖_i,x,–y and so on.

Looking at W₁ ∈ 𝓖_ixy, there are α, β ∈ 𝓘 ∖ {i, –i, x, –x, y, –y} such that W₁ ∈ 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 = {W₁}. Since |𝓖_iα| = 3, taking into account Table 2 and Lemma 3.3, 𝓖_iα ∖ {W₁} ⊂ {W₅, W₆, W₈} and 𝓖_iβ ∖ {W₁} ⊂ {W₅, W₆, W₈}. As |𝓖_iα ∖ {W₁}| = |𝓖_iβ ∖ {W₁}| = 2, there exists W ∈ {W₅, W₆, W₈} such that W ∈ 𝓖_iαβ, which contradicts Lemma 3.3 since W, W₁ ∈ 𝓖_iαβ. Therefore, there exists l₁ ∈ 𝓛 so that W₁ ∈ 𝓖_ixyl1. Similarly, there are l₂, l₄, l₅ ∈ 𝓛 such that W₂ ∈ 𝓖_i,x,–y,l2, W₄ ∈ 𝓖_i,–x,y,l4 and W₅ ∈ 𝓖_{i,–x,–y,l5}.

Let us consider W₃ ∈ 𝓖_ix. Having in view w₁, W₂ ∈ 𝓖_ix and Lemma 3.3, there are α, β, γ ∈ 𝓘∖{i,–i,x,–x,y,–y} so that W₃ ∈ 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 = {W₃}. Taking into account Table 2 and Lemma 3.3, we obtain: 𝓖_iα ∖ {W₃\\} ⊂ {W₄,…,W₈}; 𝓖_iβ ∖ {W₃} ⊂ {W₄,…,W₈}; 𝓖_iγ ∖ {W₃} ⊂ {W₄,…,W₈}. Since |𝓖_iα ∖ {W₃}| = |𝓖_iβ ∖ {W₃}| = |𝓖_iγ ∖ {W₃}| = 2 and |{W₄,…,W₈}| = 5, there exists W ∈ {W₄,…,W₈} such that W ∈ 𝓖_iεθ for ε, θ ∈ {α, β, γ}, which contradicts Lemma 3.3 since W, W₃ ∈ 𝓖_iεθ. Thus, there exists l₃ ∈ 𝓛 such that W₃ ∈ 𝓖_ixl3. Likewise, there are l₆, l₇, l₈ ∈ 𝓛 such that W₆ ∈ 𝓖_i,–x,l6, W₇ ∈ 𝓖_iyl7 and W₈ ∈ 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. □

Proposition 4.3

Ifk ∈ 𝓚, then −k ∈ 𝓚.

Proof

We are assuming |𝓖_i| = 8 for i ∈ 𝓘. The partial index distribution of the codewords W ₁, …,W₈ ∈ 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 {k₁, …, k₈} = 𝓚.

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

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 W₁, W₂, W₃ ∈ 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.

Let us now consider the codeword W₄ ∈ 𝓖_i, k_{4, −x, y}. Having in mind Lemma 3.3 we conclude that W₄ ∉ 𝓖_α, otherwise we would get W₁,W₄ ∈ 𝓖_iyα. Suppose that W₄ ∈ 𝓖_β. In these conditions, W₄, W₂ ∈ 𝓖_iβ, with W₄ ∈ 𝓖_{i,k4,−x,y,β} and W₂ ∈ 𝓖_{i,k2,x,−y,β}. Since |𝓖_iβ| = 3 (β ∈ 𝓙), there exists W ∈ 𝓖_i\{W₁, W₂, W₃, W₄} such that W ∈ 𝓖_iβ. By Table 3 we verify that W ∈ 𝓖_i,β,−x ∪ 𝓖_iβy ∪ 𝓖_i,β,−y. Consequently, taking into account W₂ and W₄, |𝓖_iβz| ≥ 2 for some z ∈ {−x, y, −y}, contradicting Lemma 3.3.

Therefore, W₄ ∈ 𝓖_γ ∪ 𝓖_δ. By a similar reasoning, we are led to the conclusion that W₅ ∈ 𝓖_γ ∪ 𝓖_δ.

We are assuming W₃ ∈ 𝓖_ik3xγδ. As k₃ ∈ 𝓚, by definition of 𝓚 we get |𝓖_ik3| = 2 . Thus, there exists k ∈ {k₁, …, k₈}\{k₃} such that k = k₃. We note that, k₃ ≠ k₁, k₂, otherwise Lemma 3.3 is contradicted. Since W₄, W₅ ∈ 𝓖_γ ∪ 𝓖_δ, taking into account Lemma 3.3 we conclude that k₃ ≠ k₄, k₅. Therefore, k ∈ {k₆, k₇, k₈}. If k₃ = k₇, then Lemma 3.3 forces W₇ ∈ 𝓖_ik7yαβ, which is a contradiction, since W₁, W₇ ∈ 𝓖_iyα. Then, k₃ ≠ k₇. By a similar reasoning we may conclude that k₃ ≠ k₈. Consequently, k₃ = k₆ and, applying once again Lemma 3.3, we must impose W₆ ∈ 𝓖_{i,k3,−x,α,β}.

Note that |𝓖_iα| = |𝓖_iβ| = 3. Since W₄, W₅ ∈ 𝓖_γ ∪ 𝓖_δ, we must obligate W₇, W₈ ∈ 𝓖_α ∪ 𝓖_β. Considering W₁ and W₂, Lemma 3.3 leads us to conclude that W₇ ∈ 𝓖_β and W₈ ∈ 𝓖_α.

Accordingly, the partial index distribution of the codewords of 𝓖_i satisfies:

Note that, as |𝓖_iγ| = |𝓖_iδ| = 3, the four empty entries of this table must be filled in with γ and δ. Thus, W₄, W₅, W₇, W₈ ∈ 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 k₁ = k₅, k₂ = k₄ and k₇ = k₈. That is, 𝓚 = {k₁, k₂, k₃, k₇} and the codewords of 𝓖_i are characterize as it is presented in Table 5.

We intend to show that if k ∈ 𝓚, then −k ∈ 𝓚. Let us focus our attention on k₃ ∈ 𝓚. We have concluded before that W₃, W₆ ∈ 𝓖_ik3, with W₃ ∈ 𝓖_ik3xγδ and W₆ ∈ 𝓖_{i,k3,−x,α,β}. In these conditions, −k₃ ∈ 𝓘 \({i,−i,x, −x,y,−y} ∪ 𝓝). That is, −k₃ ∈ 𝓘\({i,−i} ∪ 𝓙). Since 𝓘 = {i, −i} ∪ 𝓙 ∪ 𝓚, then −k₃ ∈ 𝓚.

Looking at the codewords W₇, W₈ ∈ 𝓖_ik7, we get W₇ ∈ 𝓖_γ and W₈ ∈ 𝓖_δ, or, W₇ ∈ 𝓖_δ and W₈ ∈ 𝓖_γ. In both cases −k₇ ∈ 𝓘\({i, −i} ∪ 𝓙), accordingly −k₇ ∈ 𝓚.

Now, 𝓚 = {k₁, k₂, k₃, k₇} and −k₃,-k₇ ∈ 𝓚. Either k₃ ≠−k₇ or k₃ = −k₇.

If k₃ ≠ −k₇, then −k ∈ 𝓚 for all k ∈ 𝓚.

If k₃ = −k₇ and k₁ = −k₂, then −k ∈ 𝓚 for all k ∈ 𝓚.

Assume that k₃ = −k₇ and k₁ ≠ −k₂. By this assumption it follows that −k₁, −k₂ ∈ 𝓝 = {α, β, γ, δ}. Thus, there are ε₁, ε₂ ∈ 𝓝 so that −k₁ = ε₁, −k₂ = ε₂ and the remaining elements of 𝓝, ε₃ and ε₄, satisfy ε₃ = −ε₄. As W₁ ∈ 𝓖_ik1xyα, then −k₁ ∈ {β, γ, δ}. On the other hand, since W₂ ∈ 𝓖_{i,k2,x,−y,β}, then −k₂ ∈ {α,γ,δ}. We note that, as k₁ ≠ k₂, then −k₁ ≠ −k₂.

If −k₁ = β and −k₂ = α, then γ = −δ, which is a contradiction since W₃ ∈ 𝓖_ik3xγδ.

If −k₁ = β and −k₂ = γ, then α = −δ. Analyzing Table 5 and taking into account that W₄ ∈ 𝓖_γ ∪ 𝓖_δ, we conclude that W₄ ∈ 𝓖_{i,k2,−x,y,δ}. Consequently, having in mind Lemma 3.3, W₅ ∈ 𝓖_{i,k1,−x,−y,γ}, W₇ ∈ 𝓖_ik7yβγ and W₈ ∈ 𝓖_{i,k7,−y,α,δ}, which is not possible since we are supposing α = −δ.

If −k₁ = β and −k₂ = δ, then α = −γ. Consequently, W₈ ∈ 𝓖_{i,k7,−y,α,δ}, W₇ ∈ 𝓖_ik7yβγ and W₄ ∈ 𝓖_{i,k2,−x,y,δ}. We get a contradiction since, by hypothesis, −k₂ = δ.

Combining all possibilities for −k₁ ∈ {β, γ, δ} and −k₂ ∈ {α, γ, δ}, by a similar reasoning we get always a contradiction. Therefore, −k ∈ 𝓚 for all k ∈ 𝓚.  □

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, ±1³], there exist $u₁, u₂, u₃ ∈ 𝓘 {i, −i}, with |u₁|, |u₂| and |u₃| pairwise distinct, such that U ∈ 𝓕_iu1u2u3.

By Proposition 4.2, 𝓘\{i, −i} = 𝓙 ∪ 𝓚, therefore u₁, u₂, u₃ ∈ 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, u₁, u₂, u₃ ∈ 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 u₁, u₂, u₃ ∈ 𝓚 satisfying |u₁|, |u₂| and |u₃| pairwise distinct, contradicting our assumption.□

Proposition 4.5

For allj ∈ 𝓙, |𝓓_ij ∪ 𝓔_ij| = 1. For allk ∈ 𝓚, |𝓓_ik ∪ 𝓔_ik| = 4. Furthermore, ifk ∈ 𝓚, the codewordsU₁, U₂, U₃, U₄ ∈ 𝓓_ik ∪ 𝓔_ikare such thatU₁ ∈ 𝓓_iku1 ∪ 𝓔_iku1, U₂ ∈ 𝓓_iku2 ∪ 𝓔_iku2, U₃ ∈ 𝓓_iku3 ∪ 𝓔_iku3 and U₄ ∈ 𝓓_iku4 ∪ 𝓔_iku4, with u₁, u₂ ∈ 𝓙, u₁ ≠ u₂, andu₃, u₄ ∈ 𝓚 {k, −k}, withu₃ = −u₄.

Proof

From Proposition 3.5 we get

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 V₁, V₂ ∈ 𝓖_ik and U₁, …, U₄ ∈ 𝓓_ik ∪ 𝓔_ik. We note that, the codewords of 𝓓 are of type [±3,±1²] and the codewords of 𝓔 are of type [±2², ±1]. Thus, there are v₁, …, v₆, u₁, …, u₄ in 𝓘 {i, −i, k, −k} such that:

It should be pointed out that, by Lemma 3.3, v₁, …, v₆, u₁, …, u₄ must be pairwise distinct. Therefore, {v₁, …, v₆, u₁, …, u₄} = 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, {v₁, …, v₆, u₁, …, u₄} = 𝓙 ∪ 𝓚 {k, −k}. Since V₁,V₂ ∈ 𝓖_ik, with k ∈ 𝓚, taking into account Proposition 4.2 we must impose {v₁, …, v₆} ⊂ 𝓙. Consequently, without loss of generality, u₁, u₂ ∈ 𝓙 and u₃, u₄ ∈ 𝓚 {k, −k}. Considering Proposition 4.2 we conclude that u₃ = −u₄. □

Theorem 4.6

For anyi ∈ 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 U₁, …, U₄ ∈ 𝓓_ik ∪ 𝓔_ik whose index distribution satisfies the conditions presented in Table 7, where u, −u ∈ 𝓚\{k, −k} and j₁, j₂ ∈ 𝓙, with j₁ ≠ j₂. We note that, in these conditions, 𝓚 = {k, −k, u, −u}.

Let us denote by 𝓗 the set of words of type [±2, ±1]. Consider the words P₁, P₂ ∈ 𝓗 such that P₁ ∈ Hi(2)∩Hj1(1)

and P₂ ∈ Hi(2)∩Hj2(1) . The index distribution of the codewords of 𝓓_ik ∪ 𝓔_ik and the index value distribution of the words P₁ and P₂ are represented in the following table:

By definition of perfect 2-error correcting Lee code, for each P ∈ {P₁, P₂} 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_qHi(2)∩Hjq, with j_q ∈ 𝓘\{i, −i}, is covered by a unique codeword

Thus, we may consider U₃ and U₄ as possible codewords to cover P₁ and P₂, respectively.

Suppose that P₁ is covered by U₃ and P₂ is covered by U₄. Then, we must impose

and

which contradicts Lemma 3.2, since U₃,U₄ ∈ (Di(3)∩Dk(1))∪(Ei(2)∩Ek).

Therefore, either P₁ is not covered by U₃ or P₂ is not covered by U₄.

Without loss of generality, let us assume that P₁ is not covered by U₃. Note that, U₃ ∈ 𝓓_ikj1 ∪ 𝓔_ikj1. As j₁ ∈ 𝓙, by Proposition 4.5 we get |𝓓_ij1 ∪ 𝓔_ij1| = 1. Consequently, 𝓓_ij1 ∪ 𝓔_ij1 = {U₃. Since we are assuming that U₃ does not cover P₁, considering (3), P₁ is covered by a codeword V₁ satisfying V₁ ∈ (Bi(4)∩Bj1(1))∪Cij1.

Next, we will analyze, separately, the hypotheses:

1. V₁ ∈ Bi(4)∩Bj1(1);

2. V₁ ∈ 𝓒_ij1.

1. Assume that P₁ is covered by V₁ ∈ Bi(4)∩Bj1(1).

   Assuming that P₁ is covered by V₁ ∈ Bi(4)∩Bj1(1), by Lemma 3.1 we conclude |Bi(4)∖{V1}∪Ci(3)∪Di(3)|=0. Consequently, if U ∈ {U₁,…,U₄ is such that U ∈ 𝓓, then U∈Di(1). Furthermore, P₂ must be covered by

   V2∈(Ci(2)∩Cj2(3))∪(Ei(2)∩Ej2).

   If V₂ ∈ Ei(2) ∩ 𝓔_j2, since j₂ ∈ 𝓙 we conclude, by Proposition 4.5, that V₂ = U₄. Having in mind U₁, U₂ and U₃, see Table 8, if U ∈ {U₁, U₂, U₃} is such that U ∈ 𝓔, then U ∈ Ei(1), otherwise, U, U₄ ∈ Ei(2) ∩ 𝓔_k, contradicting Lemma 3.2. Therefore, since we have concluded before that {U₁, U₂, U₃} ∩ Di(3) = ∅, we get U₁, U₂, U₃ ∈ Di(1)∪Ei(1). Taking into account the index distribution of U₁ and U₂, we must have U1∈Du(3)orU2∈D−u(3), otherwise we get U₁, U₂ ∈ (Di(1)∩Dk(3))∪(Ei(1)∩Ek(2)), contradicting, once again, Lemma 3.2.

   If V2∈Ci(2)∩Cj2(3), to avoid the contradiction of Lemma 3.2 we must impose U4∈Dk(3). Consequently, considering again Lemma 3.2, U₁, U₂, U₃ ∈ Dk(1)∪Ek(1). We recall that, we have seen before that {U₁, U₂, U₃} ∩ Di(3)=∅. Thus, in these conditions, U1∈Du(3)orU2∈D−u(3), otherwise, U₁, U₂ ∈ Ei(2)∩Ek(1), contradicting again Lemma 3.2.

   Therefore, in both cases, supposing V₂ ∈ Ei(2)∩Ej2orV2∈Ci(2)∩Cj2(3), we conclude that U1∈Du(3) or U2∈D−u(3).

   Suppose, without loss of generality, that U1∈Du(3). As u ∈ 𝓚, by Proposition 4.5 there are U₅, U₆ ∈ 𝓓_iu∪𝓔_iu satisfying U₅ ∈ 𝓓_iuj3 ∪ 𝓔_iuj3 and U₆ ∈ 𝓓_iuj4 ∪ 𝓔_iuj4, with j₃, j₄ ∈ 𝓙 distinct. Note that, j₁,…,j₄ ∈ 𝓙 are pairwise distinct, since by Proposition 4.5 we have |𝓓_ij ∪ 𝓔_ij| = 1 for all j ∈ 𝓙.

   Let us consider P3∈Hi(2)∩Hj3(1)andP4∈Hi(2)∩Hj4(1).Table 9 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].

   	i	k	u	–u	j1	j2	j3	j4
   U1	±1	±1	±3
   U2	x	x		x
   U3	x	x			x
   U4	x	x				x
   P1	±2				±1
   P2	±2					±1
   V1	±4				±1
   U5	x		x				x
   U6	x		x					x
   P3	±2						±1
   P4	±2							±1

   Taking into account the words P₃ and P₄ we may conclude, as we have concluded before for P₁ and P₂, that either P₃ is not covered by U₅ or P₄ is not covered by U₆. In fact, if U₅ covers P₃ and U₆ covers P₄, then U₅, U₆ ∈ (Di(3)∩Du(1))∪(Ei(2)∩Eu), contradicting Lemma 3.2. Let us assume, without loss of generality, that P₃ is not covered by U₅. By Proposition 4.5 it follows that |𝓓_ij3 ∪ 𝓔_ij3| = 1. Consequently, 𝓓_ij3 ∪ 𝓔_ij3 = {U₅}. As a consequence of the assumption V₁ ∈ Bi(4)∩Bj1(1) we get |Bi(4)∖{V1}∪Ci(3)∪Di(3)|=0. Thus, under these conditions and taking into account (3), P₃ must be covered by a codeword V₃ satisfying V₃ ∈ Ci(2)∩Cj3(3). Consequently, U5∈Du(3), otherwise, U₅ ∈ (Di(1)∩Dj3(3))∪(Ei(2)∩Ej3)∪(Ei∩Ej3(2)) and contradicts with the codeword V₃Lemma 3.2. However, U₁, U5∈Du(3), contradicting Lemma 3.1.

   Accordingly, P₁ can not be covered by the codeword V₁ ∈ Bi(4)∩Bj1(1).

2. Assume thatP₁is covered byV₁ ∈ 𝓒_ij1.

   Since V₁ ∈ 𝓒, then V₁ is a codeword of type [±3, ±2]. According with what is being supposed, V₁ ∈ Ci(3)∩Cj1(2)orV1∈Ci(2)∩Cj1(3). Consider U₃ ∈ 𝓓_ikj1 ∪ 𝓔_ikj1. In order to have Lemma 3.2 fulfilled we must force U₃ ∈ Di(1)∩Dk(3)∩Dj1(1). Schematically, we get Table 10.

   Table 10

   Index distribution on 𝓒_i ∪ 𝓓_i ∪ 𝓔_i and on 2 words of type [±2, ±1].

   	i	k	u	–u	j1	j2
   U1	x	x	x
   U2	x	x		x
   U3	±1	±3			±1
   U4	x	x				x
   P1	±2				±1
   P2	±2					±1
   V1	x				x

   Taking into account U₃, by Lemma 3.2 we must have U₁, U₂, U₄ ∈ Dk(1)∪Ek(1). Besides, U1∈Du(3)orU2∈D−u(3), otherwise, U₁, U₂ ∈ (Di(3)∩Dk(1))∪(Ei(2)∩Ek(1)), contradicting Lemma 3.2.

   Let us assume, without loss of generality, that U1∈Du(3),

   Proceeding as in the previous case, we will consider U₅ ∈ 𝓓_iuj3 ∪ 𝓔_iuj3 and U₆ ∈ 𝓓_iuj4 ∪ 𝓔_iuj4, with j₃, j₄ ∈ 𝓙 and distinct. We will consider also P₃ ∈ Hi(2)∩Hj3(1) and P₄ ∈ Hi(2)∩Hj4(1). Gathering the information obtained so far, one has the index distribution presented in Table 11.

   Table 11

   Index distribution on 𝓒_i ∪ 𝓓_i ∪ 𝓔_i and on 4 words of type [±2, ±1].

   	i	k	u	–u	j1	j2	j3	j4
   U1	±1	±1	±3
   U2	x	x		x
   U3	±1	±3			±1
   U4	x	x				x
   P1	±2				±1
   P2	±2					±1
   V1	x				x
   U5	x		x				x
   U6	x		x					x
   P3	±2						±1
   P4	±2							±1

   As seen in the previous case, either U₅ does not cover P₃ or U₆ does not cover P₄. Assume, without loss of generality, that P₃ is not covered by U₅. By Proposition 4.5 we get 𝓓_ij3 ∪ 𝓔_ij3 = {U₅}. Therefore, considering (3), P₃ must be covered by a codeword V₃ ∈ (Bi(4)∩Bj3(1)) ∪ 𝓒_ij3. If V₃ ∈ 𝓒_ij3, then, by Lemma 3.2, we must impose U₅ ∈ Du(3) and, consequently, | Du(3)| ≥ 2, contradicting Lemma 3.1. Accordingly, V₃ ∈ (Bi(4)∩Bj3(1)).

   Taking into account Lemma 3.1, |Bi(4)∖{V3}∪Ci(3)∪Di(3)|=0. Thus, by (3) we may conclude that P₄ must be covered by a codeword

   V4∈(Ci(2)∩Cj4(3))∪(Ei(2)∩Ej4).

   Note that, if V4∈Ci(2)∩Cj4(3), then, by Lemma 3.2, U₆ ∈ Du(3) implying | Du(3)| ≥ 2 and contradicting Lemma 3.1. Thus, V₄ ∈ Ei(2)∩Ej4. By Proposition 4.5, |𝓓_ij4 ∪ 𝓔_ij4| = 1 leading to 𝓓_ij4 ∪ 𝓔_ij4 = {U₆ and, consequently, V₄ = U₆. Since U₁ ∈ Di(1)∩Dk(1)∩Du(3), taking into account Lemma 3.2, we must force U₆ ∈ Ei(2)∩Eu(1)∩Ej4(2). 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.

   Table 12

   Index distribution on 𝓑_i ∪ 𝓒_i ∪ 𝓓_i ∪ 𝓔_i and on 4 words of type [±2, ±1].

   	i	k	u	–u	j1	j2	j3	j4
   U1	±1	±1	±3
   U2	x	x		x
   U3	±1	±3			±1
   U4	x	x				x
   P1	±2				±1
   P2	±2					±1
   V1	x				x
   U5	x		x				x
   U6	±2		±1					±2
   P3	±2						±1
   P4	±2							±1
   V3	±4						±1

   Let us now focus our attention on –u ∈ 𝓚. By Proposition 4.5, there are codewords U₇, U₈ ∈ 𝓓_i,–u ∪ 𝓔_i,–u, so that, U₇ ∈ 𝓓_i,–u,j5 ∪ 𝓔_i,–u,j5 and U₈ ∈ 𝓓_i,–u,j6 ∪ 𝓔_i,–u,j6, with j₅, j₆ ∈ 𝓙 distinct. Note that, by Proposition 4.5, |𝓓_ij ∪ 𝓔_ij| = 1 for all j ∈ 𝓙, and so j₁,…,j₆ are pairwise distinct. Taking into account the existence of the words P₅ ∈ Hi(2)∩Hj5(1) and P₆ ∈ Hi(2)∩Hj6(1), we obtain the index distribution presented schematically in Table 13.

   Table 13

   Index distribution on 𝓑_i ∪ 𝓒_i ∪ 𝓓_i ∪ 𝓔_i and on 6 words of type [±2, ±1].

   	i	k	u	–u	j1	j2	j3	j4	j5	j6
   U1	±1	±1	±1
   U2	x	x		x
   U3	±1	±1			±1
   U4	x	x				x
   P1	±2				±1
   P2	±2					±1
   V1	x				x
   U5	x		x				x
   U6	±2		±1					±2
   P3	±2						±1
   P4	±2							±1
   V3	±4						±1
   U7	x			x					x
   U8	x			x						x
   P5	±2								±1
   P6	±2									±1

   By a similar reasoning to the one done with the words P₁, P₂, P₃, P₄ ∈ Hi(2), we conclude that either P₅ is not covered by U₇ or P₆ is not covered by U₈. Let us assume, without loss of generality, that U₇ does not cover P₅. Then, considering (3) we are lead to conclude that P₅ must be covered by a codeword

   P5∈(Bi(4)∩Bj5(1))∪(Cij5).

   As V₃ ∈ Bi(4)∩Bj3(1), by Lemma 3.1, P₅ ∈ Ci(2)∩Cj5(3). Consequently, taking into account Lemma 3.2, we must force U7∈D−u(3).

   Focus our attention on the codeword U₂ ∈ 𝓓_i,k,–u ∪ 𝓔_i,k,–u. Having in mind the index value distribution of the codewords V₃, U₃ and U₇ and considering Lemma 3.1, we conclude that U₂ ∈ 𝓔_i. Consequently, either U₂ ∈ 𝓔_i ∩ Ek(2) or U₂ ∈ 𝓔_i ∩ E−u(2). If U₂ ∈ 𝓔_i ∩ Ek(2), then the index value distribution of U₂ and U₃ contradicts Lemma 3.2. If U₂ ∈ 𝓔_i ∩ E−u(2), the index value distribution of U₂ and U₇ contradicts also Lemma 3.2.

   In both hypotheses, P₁ covered by V₁ ∈ Bi(4)∩Bj1(1) or P₁ covered by V₁ ∈ 𝓒_ij1, we get a contradiction.  □