By Xu B.G.

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**Extra info for A 3-color Theorem on Plane Graphs without 5-circuits**

**Example text**

Step 1 There exists no NITA such that e7 ∈ Supp{z12 }. Suppose that e7 ∈ Supp{z12 }. Since e7 ∈ Supp{d8 }, there exist α3 ∈ Supp{z12 }, g4 ∈ Supp{d8 } or α4 ∈ Supp{z12 }, g3 ∈ Supp{d8 } such that e7 = α3 + g4 or α4 + g3 . Substep 1 There exists no NITA such that e7 = α3 + g4 . If e7 = α3 + g4 , then there exist z9 ∈ N ∗ B and h4 ∈ B such that z12 = α3 + z9 and d8 = g4 + h4 . Furthermore, b¯4 b5 = α3 + z9 + b8 , b¯3 b9 = c7 + b8 + α¯ 3 + z¯ 9 , b¯3 x5 = b8 + α3 + g4 , Hence b3 α3 = x5 + i4 , i4 ∈ B, b3 α¯ 3 = b9 , which is impossible for (b3 α¯ 3 , b3 α¯ 3 ) = 1 and (b3 α3 , b3 α3 ) = 2.

Here m9 ∈ B, otherwise l¯4 l4 will contains two b8 , which is a contradiction. Since b3 (b3 b¯3 ) = 2b3 + b9 + x5 + y7 , (b3 g3 )g¯3 = g¯ 3 x5 + g¯3 l4 , it follows that g¯ 3 l4 = b3 + b9 and g¯ 3 x5 = b3 + x5 + y7 = b¯3 b5 . 25) We assert that Supp{n8 } ∩ {b3 , x5 , y7 , l4 } = ∅. Obviously y7 ∈ Supp{n8 }. By the expressions of b3 h5 and b3 b¯3 , we have that b3 ∈ Supp{n8 }. If x5 ∈ Supp{n8 }, then h5 ∈ Supp{b¯3 x5 }, a contradiction. If l4 ∈ Supp{n8 }, then h5 ∈ Supp{b¯3 l4 }, a contradiction.

41) Now we assert that c5 and e5 are nonreal. 41). So 1 ∈ Supp{b3 a5 }, a contradiction. Hence c5 and a5 are nonreal, and c5 = a¯ 5 by the expression b82 . Hence b3 d¯6 = d8 + w5 + c5 . 38), we have that (b3 c¯5 , d6 ) = (b3 c¯5 , s4 ) = 1, which implies that (b3 c¯5 , b3 c¯5 ) = 3. Furthermore (b3 c5 , b3 c5 ) = 3. Let b3 c5 = v8 + δ3 + λ4 . Thus we may set b¯3 δ3 = c5 + Δ4 . Then δ¯3 δ3 = 1 + b8 , and b3 (δ3 δ¯3 ) = b3 + b3 b8 = 2b3 + b9 + x5 + y7 , (b3 δ¯3 )δ3 = c¯5 δ3 + Δ¯ 4 δ3 . We have that Δ¯ 4 δ3 = b3 + b9 and c¯5 δ3 = b3 + x5 + y7 .