Proof.  Let  and  be the bipartition of vertex set  of a tree T, where  and .
First, notice that  is a set-ordered graceful labeling of T, then  for  and  for , and each edge  has its label .
Another labeling 
 is defined for the tree 
T as: 
 for 
, and
        
        for 
. The vertex set and edge set is
        
        then we obtain:
(i) .
(ii) Each edge 
 with another edge 
 corresponds and holds 
 such that
        
(iii) Let 
 for 
, so
        
        then we obtain a set 
, where 
p is even, or a set 
, where 
p is odd. Therefore, each edge 
 and another edge 
 correspond, so 
, except that edge 
e holds 
 as 
p is even.
(iv) 
 from Equation (
4).
(v)  for each edge  corresponding to one vertex w, and  for each vertex z corresponding to one edge , except the singularity .
(vi) We obtain  for the bipartition  of .
Therefore,  admits a 6C-labeling.
For the converse, let 
 be a 6C-labeling of 
T. According to the property (iv) and 
, we obtain the edge label and the vertices’ label set 
 and 
, respectively. A labeling 
 is defined as: 
 for 
, which gives 
; 
 for each edge 
, so 
. The property (i) enables us to compute
        
        that is 
 is graceful. The graceful labeling 
 is set-ordered according to the property (vi).    □
 Proof.  Let each tree  of p vertices admit a set-ordered graceful labeling  and  be the bipartition of  with . Therefore, we have  where  and  for  with . Then, we can label the vertices’ set as  for ,  for  and label the edge set as  for each edge , and  for .
We define another labeling  of  as:  for  and  for each edge . Therefore, we can compute  and .
Next, we define another labeling  of  as:  for each vertex  and  for each edge . Thereby, we obtain , .
Notice that  and . By Lemma 1, we have proven the theorem.    □
 Proof.  According to the supposition of the corollary, let a tree T have its own vertex bipartition  with  and  with  and . For a set-ordered graceful labeling f of T, we obtain  for  and  for , and  for each edge :
(1) The labeling 
 is defined for a copy 
 of 
T with 
 as: 
 for 
 and 
 for 
; immediately,
        
Therefore,  is an ogl of , since  is an even-set,  is an odd-set, and  is an odd-set as well. Next, we another another graph  obtained by copying T with  and make a complementary labeling of the ogl by setting  for ; clearly, . Moreover, , ; we can see  and . Therefore,  is the complementary of . Therefore,  admits a togl.
The proof is finished for the corollary.    □
 Proof.  For each , each tree  has  vertices and bipartition , where  and  with .
By the assumption of the theorem, each tree  for  has a set-ordered graceful coloring  holding , , , as well as , which shows that  and .  has a graceful labeling  defined as  for  and  such that  and .
We join the vertex  with the vertex  by an edge, ; we join the vertex  with the vertex  by an edge, so the resulting tree is denoted as T. Next, we define a labeling g of T in the following steps. Let  and , .
We define another labeling g as follows:
Step 1. For each , the vertices’ color of  as , , where .
Step 2. The vertices’ color of  as , , where .
Step 3. The vertices’ color of  as  for each , , where .
Step 4. The edges  are colored as  for , , .
Step 5. The edges  are colored as  for , .
We can verify the vertices’ and edges’ set as follows:
;
;
;
;
 for .
Let us continue to validate the restriction of labeling g as follows:
(i) Each edge 
:
        
        holds true.
(ii) Each edge 
 corresponding to another edge 
 holds 
 such that
        
(iii) Let 
 for 
, so
        
        which distributes a set 
 if 
p is even or a set 
 if 
p is odd. Thereby, each edge 
 corresponds to another edge 
 such that 
, except that edge 
e holds 
 as 
p is even.
The proof of (iv)–(vi) is the same as Lemma 1.
Hence, we claim that the labeling f admits really a 6C-labeling defined in Definition 2.
Thereby, we claim that 
 and 
, which shows that 
g is a 6C-labeling of 
T. See an example in 
Figure 4 and 
Figure 5.    □