By Jurisic A.

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**Example text**

Since b¯3 (b3 α4 ) = b¯3 y8 + b3 γ4 = b8 + α4 + β8 + x¯4 + b3 γ4 , (b¯3 b3 )α4 = α4 + α4 b8 = 2α4 + c7 + b8 + β8 + e5 , we have that x¯4 + b3 γ4 = α4 + c7 + e5 , which implies that x¯ 4 = α4 . 21) b3 x4 = j8 + x4 , j8 ∈ B and b3 α¯ 4 = b9 + r3 , a contradiction. Step 3 There exists no NITA such that z12 = α5 + β7 . If z12 = α5 + β7 , then b3 (α¯ 5 + β¯7 ) = 2b9 + r3 + t4 + 2b9 + x11 . We need a sum of degree 6 to make up b3 α¯ 5 for b9 ∈ Supp{b3 α¯ 5 }. If r3 ∈ Supp{b3 β¯7 }, then b¯3 r3 contains an element of degree 2, a contradiction.

3 and h6 , i6 ∈ B, a contradiction. Step 3 There exists no NITA such that d8 = c4 + d4 . 51) b82 = 1 + c7 + 3b8 + c4 + h6 + d4 + i6 + h¯ 6 + i¯6 . 50) and (b¯4 b5 , b¯4 b5 ) = 3, we have that i6 , h6 ∈ B. 53) b3 d4 = y6 + p6 , p6 ∈ B. 54) Substep 1 h6 and i6 are nonreal. If h6 is real, then there exist e3 and f3 such that b¯ 3 h6 = b¯9 + x¯6 + e3 , b3 e3 = f3 + h6 . 2. Since 46 2 Splitting of the Main Problem into Four Sub-cases b3 (e3 e¯3 ) = 2b3 + b9 + x6 + y6 , (b3 e3 )e¯3 = h6 e¯3 + f3 e¯3 , hence f3 e¯3 = b3 + x6 or b3 + y6 , h6 e¯3 = b3 + b9 + y6 or b3 + b9 + x6 .

Since b¯ 3 x4 = b8 + x4 , we may set b3 x¯4 = a3 + b9 and b3 x4 = x4 + e8 , for some a3 , e8 ∈ B. Suppose that b3 a¯ 3 = x4 + l5 , for some l5 ∈ B. Then b3 (b¯3 x4 ) = b3 b¯9 + b3 a¯ 3 = c7 + b8 + x4 + s8 + x4 + l5 , b¯3 (b3 x4 ) = b¯3 x4 + b¯3 e8 = b8 + x4 + b¯3 e8 . Hence b¯3 e8 = c7 + s8 + x4 + l5 . Therefore e8 ∈ Supp{b3 c7 }. Let b3 c7 = b9 + e8 + e4 , e4 ∈ B. By (b3 b¯3 )a¯ 3 = b¯3 (b3 a¯ 3 ), we obtain a¯ 3 b8 = b¯ 9 + b¯3 l5 . Since (a¯ 3 b8 , a¯ 3 b8 ) = (a3 a¯ 3 , b82 ) = (b3 b¯3 , b82 ) = 4, we have that (b¯3 l5 , b¯3 l5 ) = 3.