3. Proof of Theorem 3
Let  (, , respectively) represent the set of vertices in G of degree i (at most i, at least i, respectively). For the graph G, a graph  is smaller than it if one of the below conditions is satisfied.
	  
- (i)
- ; 
- (ii)
- In the lexicographic order,  precedes , where . 
We say a graph is minimal for some property if no smaller graph admits the property. To obtain a smaller graph, , than G, we will frequently delete some edges or split some vertices of G in the following.
From now on, let 
G be a minimal graph not satisfying Theorem 3 and 
. Then, any graph 
 that is smaller than 
G has a tnsd 
k-coloring 
. By splitting a vertex 
v into two vertices 
 and 
, we obtain a smaller graph 
 than 
G, then 
 has a tnsd 
k-coloring 
. If we can obtain a tnsd 
k-coloring of 
G from the 
 by sticking 
 and 
 into 
v, we say that 
 and 
 can be 
properly stunk. In the following, we will often extend a tnsd 
k-coloring 
 of 
 to a tnsd 
k-coloring 
 of 
G to obtain a contradiction. Set 
. Assume that 
X is a subset of 
 and 
 is a mapping. For each 
, set
      
Assume that u is a -vertex in G and  is a mapping satisfying the conditions as follows:
	  
- (i)
-  for any adjacent or incident elements ; 
- (ii)
-  for any two adjacent vertices . 
Since , . Thus, we can extend the  to a tnsd k-coloring  of G, which is a contradiction. For simplicity, we will omit the colors of all -vertices in the following discussion.
In 
Figure 1 and 
Figure 2, a black vertex represents that its neighbors are all depicted and a white vertex means that its neighbors may not be all depicted.
Claim 1. The graph G contains no configuration in Figure 1.  Proof.  Suppose that 
G contains the configuration in 
Figure 1. By splitting the two 3-vertices 
 of the configuration in 
Figure 1 into 
 and 
, respectively, we obtain a smaller graph 
 than 
G. Then, 
 has a tnsd 
k-coloring 
. Without loss of generality, set 
, 
, 
, 
, 
, 
, 
, 
, and 
 (see 
Figure 2).
If we can properly stick  with , respectively, we can obtain a tnsd k-coloring  of G, which is a contradiction. Thus, assume that at least one of  and  cannot be stuck properly below. Let .
Case 1. Suppose that one of  and  cannot be stuck properly. If  can not be stuck properly, . Without loss of generosity, set .
Set , , and . Then, we assign , , and  for every .
Set , , and . Then, we assign , , , , and  for every .
Set . If , we assign , , , and  for every . If  and , we assign , , , , , and  for every . If  and , we assign , , , , and  for every .
Set . If  and , we assign , , , and  for every . If  and , we assign , , , , , , and  for every . If , we assign , , , , , and  for every .
In each case above, by sticking  and  into  for , we obtain the graph G. Since both  and  are 3-vertices in G, we can obtain a tnsd k-coloring  of G from . By symmetry, if  cannot be stuck properly, we can also obtain a tnsd k-coloring  of G. It is a contradiction.
Case 2. Assume that both  and  cannot be stuck properly. Then,  and . Let , , , , , , and  for every . By sticking  and  into  for , we obtain the graph G. Since both  and  are 3-vertices in G, we can obtain a tnsd k-coloring  of G from , which is a contradiction. □
 The below Claim 2 follows from (3) in Theorem 2.
Claim 3. If , then 
 Proof.  By contradiction, suppose that there is a vertex 
 and a vertex 
 such that 
. Let 
. Then, 
 has a tnsd 
k-coloring 
. We first erase the colors on 
u and 
v from 
. Without loss of generality, set 
 and 
. Then,
        
Let 
 be a map on 
 obtained from 
 by assigning 
, 
, 
, and 
 for every 
. Set
        
With the help of MATHEMATICA, we have . By Theorem 4, there is a color , a color , and a color  such that . Therefore, we can extend the  to a tnsd k-coloring  of G by recoloring  with colors  and coloring  with color , respectively. It is a contradiction.    □
 Claim 4. If ,  and .
 Proof.  We first prove 
 when 
. By contradiction, suppose that there is a vertex 
 and a vertex 
 such that 
. Let 
. Then, 
 has a tnsd 
k-coloring 
. We first erase the colors on 
u and 
v from 
. Note that 
v is a 
-vertex. Then,
        
Let 
 be a map on 
 obtained from 
 by assigning 
, 
, and 
 for every 
. Set
        
By applying MATHEMATICA, we have . By Theorem 4, there is a color  and a color  such that .
Since v is a -vertex, we can obtain a tnsd k-coloring of G from the  by recoloring u with color  and coloring  with color , respectively. This is a contradiction.
Next, we show 
 when 
. Suppose, to the contrary, that there is a vertex 
 and two vertices 
 such that 
. Let 
. Then, 
 has a tnsd 
k-coloring 
. We first erase the colors on 
u, 
, and 
 from 
. Then,
        
Let 
 be a map on 
 obtained from 
 by assigning 
, 
, and 
 for 
, and 
 for every 
. Set
        
With the help of MATHEMATICA, we have . By Theorem 4, there is a color , a color , and a color  for  such that . Therefore, we can extend the  to a tnsd k-coloring  of G by recoloring  with colors  and coloring  with color  for , respectively. This is a contradiction.    □
 Claim 5. If  with , then,
- (1)
 
- (2)
-  when  
 Proof.  Suppose to be contrary that there is a vertex 
, a vertex 
, and 
 vertices 
 with 
 such that 
 and 
 for 
. Let 
. Then, 
 has a tnsd 
k-coloring 
. We first erase the colors on 
u and 
 with 
 from 
. Without loss of generality, set 
 with 
. Note that 
 is a 
-vertex. Then,
        
Let 
 be a map on 
 obtained from 
 by assigning
        
        and 
 for every 
 Set
        
When , , and we have  by applying MATHEMATICA. When , , and we have  by applying MATHEMATICA. By Theorem 4, there is a color , a color  with , and a color  with  such that  when . Since  is a -vertex, we can extend the  to a tnsd k-coloring of G by recoloring  with colors  and coloring  with colors , respectively. This is a contradiction.    □
 Claim 6. If  with , then
- (1)
 
- (2)
-  when  
- (3)
-  when  and  
 Proof.  By contradiction, suppose that there is a vertex 
, a vertex 
, a vertex 
, 
 vertices 
 with 
 such that 
, 
, and 
 for 
. Let 
. Then, 
 has a tnsd 
k-coloring 
. We first erase the colors on 
u and 
 with 
 from 
. Without loss of generality, set 
. Note that 
 is a 
-vertex and 
 is a 
-vertex with 
. Then,
        
Let 
 be a map on 
 obtained from 
 by assigning
        
        and 
 for every 
 Set
        
When , , and we have  with the help of MATHEMATICA. When , , and we have  with the help of MATHEMATICA. By Theorem 4, there is a color , a color , and a color  with  such that  when . Since  is a -vertex with , we can obtain a tnsd k-coloring of G from the  by recoloring  with colors  and coloring  with colors , respectively. This is a contradiction.    □
 Claim 7. If , then
(1)
(2) when ,
(3) when  and ,
(4) when ,  and .
 Proof.  Suppose to the contrary that there is a vertex 
, a vertex 
, a vertex 
, a vertex 
, and three vertices 
 such that 
, 
, 
, and 
 for 
. Let 
. Then, 
 has a tnsd 
k-coloring 
. We first erase the colors on 
u and 
 with 
 from 
. Without loss of generality, set 
. Note that 
 is a 
-vertex, 
 is a 
-vertex, and 
 is a 
-vertex with 
. Then,
        
Let 
 be a map on 
 obtained from 
 by assigning
        
        and 
 for every 
 Set
        
With the help of MATHEMATICA, we have . By Theorem 4, there is a color , a color , and a color  with  such that . Since  is a -vertex with , we can obtain a tnsd k-coloring of G from the  by recoloring  with colors  and coloring  with colors , respectively. This is a contradiction.    □
 Claim 8. Let  be a 3-cycle in G. Then, .
 The following Equation (
1) follows from Claims 3∼7.
      
Let  Then,  for each .
Claim 9. If , then,
- (1)
- , 
- (2)
-  if , 
- (3)
-  if , 
- (4)
-  if  and , 
- (5)
-  if . 
 Proof.  Note that 
 by the definition of 
H and Claim 2 for every 
. By the definition of 
H, (1), (2), (3), and (4) follow from Equation (
1). Below, we prove (5).
(5) Suppose that  when . Without loss of generosity, set . Then,  by (2). Thus,  by Claim 3, and u has at least one -neighbor v in G with  by (2). Hence,  by (3) and the definition of H, which is a contradiction. Therefore, the statement (5) holds.    □
 By a similar discussion to that of Claim 9, we can obtain the following Claims 10∼12.
Claim 10. If , then .
 Claim 11. If  with , then,
- (1)
- , 
- (2)
-  when . 
 Claim 12. If  with , then,
- (1)
- , 
- (2)
-  when , 
- (3)
-  when  and . 
 By Claims 8 and 9, the below claim is immediate.
Claim 13. Let  be a 3-cycle in H. Then,
- (1)
- , 
- (2)
-  when . 
 From now on, we always assume that H has been embedded on the plane such that the number of crossings is as small as possible. Let  be the associated planar graph of H and . If  is incident with two 3-faces and u is a real ℓ-vertex in , we call u a badℓ-neighbor of v. If  is incident with exactly one 3-face and u is a real 3-vertex in , we call u a special 3-neighbor of v. Let  denote the number of bad ℓ-neighbors (special 3-neighbors) of v in . Unless stated otherwise, an ℓ-vertex represents a real ℓ-vertex in the following. A 4-face is called a good 4-face if it is incident at most a real -vertex and a bad 4-face otherwise. By Claims 1 and 9, the below Claims 14 and 15 are immediate.
Claim 14. If , then,
- (1)
- , ,  and , 
- (2)
-  when . 
 Claim 15. Every k-face is incident with at most  real -vertices in .
 Let 
 denote the number of connected components of 
. Note that each plane graph has exactly one unbounded face. Then, we have
      
      and Euler’s formula
      
Below, we will apply the discharging method [
13] on the planar graph 
 to complete the proof of Theorem 3. Set 
 for any 
 and 
 for any 
. By Euler’s formula,
      
      and
      
In order to redistribute charges on  and keep the total charges unchanged, we make some discharging rules as follows.
	  
- (R1) 
- Let  be an edge of .
	   - (R1.1) 
- If u is a bad 3-neighbor of v, u receives 1 from v. 
- (R1.2) 
- If u is a special 3-neighbor of v, u receives  from v. 
- (R1.3) 
- If u is a bad 4-neighbor of v, u receives  from v. 
- (R1.4) 
- If u is a bad 5-neighbor of v, u receives  from v. 
 
- (R2) 
- Each real ℓ-vertex with  receives 1 from every -face incident with it. 
- (R3) 
- Let  be an edge and u be a false 4-vertex of . - (R3.1) 
- If , u receives 1 when v is not adjacent to any real -vertex and  otherwise from v. 
- (R3.2) 
- If  with , u receives 2 when  and 1 otherwise from v. 
- (R3.3) 
- If f is a good 4-face or -face incident with u, u receives 1 from f. 
 
For each 
, we use 
 to represent the new charge after applying the rules. Then, one can obtain
      
Thus, there exists a vertex or a face whose charge is negative.
Firstly, it is easy to verify that the new charge of every face is non-negative by Claim 15 and the discharging rules.
Next, we show that the new charge of every real vertex is non-negative. Pick arbitrarily a real vertex v from . Note that v is adjacent to at most a false 4-vertex as G (and thus, H) is an IC-planar graph. Note also that  and  by the definition of  and Claim 9.
 For . If v is incident with three -faces, then  by (R2). If v is incident with two -faces, it is a common special 3-neighbor of two neighbors, and so  by (R1.2) and (R2). If v is incident with one -face, then it is a special 3-neighbor of one neighbor and a common bad 3-neighbor of the other two neighbors, respectively, and so  by (R1.1), (R1.2), and (R2). If v is not incident with any -faces, it is a common bad 3-neighbor of three neighbors; so,  by (R1.1).
 For . If v is incident with at least two -faces,  by (R2). If v is incident with one -face, it is a common bad 4-neighbor of two neighbors; so,  by (R1.3) and (R2). If v is not incident with any -faces, it is a common bad 4-neighbor of four neighbors; so,  by (R1.3).
 For . If v is incident with at least one -faces,  by (R2). If v is not incident with any -faces, it is a common bad 5-neighbor of five neighbors; so,  by (R1.4).
 For .  as no rule is applied to it.
 For . Note that v is adjacent to at most a real 5-vertex in H (and so, in ) by Claim 10. If v is not adjacent to any real 5-vertex in ,  by (R3.1). If v is adjacent to a real 5-vertex in ,  by (R1.4) and (R3.1).
 For  with . If ,  by (R1), (R3.2), and Claim 11. If , ; so, v is adjacent to at most  real -vertices in  by Claim 11 and the definition of . Thus,  by (R1) and (R3.2).
 For  with . If ,  by (R1), (R3.2), and Claim 12. If , ; so, v is adjacent to at most  real -vertices in  by Claim 12 and the definition of . Thus,  by (R1) and (R3.2).
 For . By Claim 14,  by (R1) and (R3.3).
Hence, the new charge of every real vertex is non-negative.
Finally, we prove the new charge of every false 4-vertex is non-negative. Choose arbitrarily a false 4-vertex u from . Note that u is adjacent to at most two -vertices in  by the definition of  and Claim 9. Assume that the edges  and  intersect at u in H. Then, . The subgraph of  induced by  is denoted by . Note that if  is a -vertex with ,  is a -vertex and  is not adjacent to any real -vertex when  is a 7-vertex in  by the definition of  and Claim 9.
 Let 
 contain no 3-face (see 
Figure 3). Then, 
u is incident with at least three faces, each of which either is a good 4-face or a 
-face since 
 by Claim 9; so, 
 by (R3.3).
 Let 
 contain exactly one 3-face (see 
Figure 4). Note that 
 and at most one of 
 and 
 is equal to 3 by Claim 9. If 
 and 
, 
 by (R3.3) since 
u is incident with at least two faces, each of which either is a good 4-face or a 
-face. If 
 is equal to 3, 
 by (R1.2) and (R3) since 
u is incident with at least two faces, each of which either is a good 4-face or a 
-face. By symmetry, 
 when 
 is equal to 3.
 Let 
 contain two 3-faces (see 
Figure 5). Note that 
 by Claim 9 and 
 of (b) in 
Figure 5 is not a 3-vertex by Lemma 1. If 
, 
 by (R3.3) since 
u is incident with two faces, each of which either is a good 4-face or a 
-face. If 
, 
 by (R1.2) and (R3.3), since 
u is incident with two faces, each of which either is a good 4-face or a 
-face. If 
, 
 by (R1.2) and (R3.3) since 
u is incident with one face, which either is a good 4-face or a 
-face.
 Let 
 contain three 3-faces (see 
Figure 6). Note that 
 by Claim 9 and neither 
 nor 
 is a 3-vertex by Lemma 1 in 
Figure 6. If 
, at least one neighbor of 
u is a 
-vertex, and when one neighbor of 
u is 7-vertex, it is not adjacent to any real 
-vertex by Claim 13; so, 
 by (R1) and (R3.3) since 
u is incident with one face, which either is a good 4-face or a 
-face. If 
 and 
, 
u is adjacent to at least two 
-vertices; so, 
 by (R1.3), (R1.4), and (R3.3) since 
u is incident with one face, which either is a good 4-face or a 
-face. If 
 and 
, 
 or 
. Since 
 and 
 are symmetrical in 
Figure 6, we can set 
. Then, 
, 
, and 
; so, 
 by (R1.2), (R3.2), and (R3.3) since 
u is incident with one face, which either is a good 4-face or a 
-face. If 
, 
, 
, 
, and 
 by Claims 9 and 10. If 
, 
 by (R3.2). If 
, then we may as well set 
 since 
 and 
 are symmetrical in 
Figure 6. Thus, 
; so, 
 by (R1.2) and (R3.2). If 
, 
. Thus, 
 and 
. So, 
 by (R1.2) and (R3.2).
 Let 
 contain four 3-faces (see 
Figure 7). Then, 
 for 
 by Lemma 1. Note that 
 by Claim 9. If 
 and 
, then 
u is adjacent to at least two 7-vertices by Claim 13. When 
u is adjacent to two 7-vertices, each of the two 7-vertices is not adjacent to any real 
-vertex by Claim 13, and so 
 by (R3.1). When 
u is adjacent to at least three 7-vertices, 
 by (R3.1). If 
 and 
, we may as well set 
 by symmetry. Then, at least one of 
, 
 and 
 is a 7-vertex by Claim 13. When one of 
, 
 and 
 is a 7-vertex, this 7-vertex is not adjacent to any real 
-vertex by Claim 13, and so 
 by (R3.1) and (R3.2). When at least two of 
, 
 and 
 are 7-vertices, 
 by (R3.1) and (R3.2). If 
 and 
, 
 by (R3.2). If 
 and 
, 
u is adjacent to three 
-vertices by Claim 9; so, 
 by (R1.3) and (R3.2). If 
 and 
, 
u is adjacent to three 
-vertices by Claim 9; so, 
 by (R1.4), (R3.1), and (R3.2).
Hence, the new charge of every false 4-vertex is non-negative.
By the above analysis, we obtain that the new charge of every member in  is non-negative. This is a contradiction. The proof of Theorem 3 is completed.