To apply theorem 1.1 in our theorem

To apply theorem 1.1 in our theorem, we must seek the degree of each vertex
of the glued graph. There are relationships between the degree of the vertex of
the glued graph, the two original graphs and the clone that can be stated in the
following lemma.
Lemma 2.1. Let G1 and G2 be any nontrivial graphs, H1 ⊆ G1, H2 ⊆ G2. Let H0
be the labeled clone of the labeled glued graph G1✁✄¯
H0
G2 and i be a natural number.
Then the degree of each vertex in the labeled glued graph is as follows:
1.) degG1✁✄¯
H0
G2
(ui
, v0) = degG1
ui for the vertex obtained from only ui of
G1H1,
2.) degG1✁✄¯
H0
G2
(u0, vi) = degG2
vi for the vertex obtained from only vi of
G2H2,
3.) degG1✁✄¯
H0
G2
(ui
, vi) = degG1
ui + degG2
vi − degH0 wi for the vertex obtained
from both ui of H1 and vi of H2.
Proof. Let G1 and G2 be any nontrivial graphs, H1 ⊆ G1, H2 ⊆ G2. Let
H0 be the labeled clone of the labeled glued graph G1✁✄¯
H0
G2 and i be a natural
number. The vertices (ui
, v0),(u0, vi) and (ui
, vi) are the vertices which are obtained
from only ui of G1H1, only vi of G2H2 and both ui of H1 and vi of
H2, respectively. Since ui
in G1H1 becomes (ui
, v0) and vi
in G2H2 becomes
(u0, vi) without combining of the two labeled original graphs, the degree is still
the same. So degG1✁✄¯
H0
G2
(ui
, v0) = degG1
ui and degG1✁✄¯
H0
G2
(u0, vi) = degG2
vi
. Finally,
we see that each edge of H0
contributes twice in the degree of (ui
, vi). Thus,
degG1✁✄¯
H0
G2
(ui
, vi) = degG1
ui + degG2
vi − degH0 wi
. ¥

ผลลัพธ์ (อังกฤษ) 2:

To Apply theorem 1.1 in our theorem, we must Seek The Vertex Degree of each
of The glued graph. There are Relationships between The Degree of The Vertex of
The glued graph, The Original Two Graphs and The Clone Can that be stated in The
following Lemma.
Lemma 2.1. Let G1 and G2 be any nontrivial graphs, H1 ⊆ G1, H2 ⊆ G2. Let H0
be labeled The Clone of The labeled graph glued G1✁✄¯
H0
G2 and I be a natural number.
Then The Degree of each Vertex in The graph is labeled glued As follows:
1.) DegG1✁✄¯
H0
G2
(ui.
, V0) = DegG1
obtained from only ui ui for The Vertex of
G1 H1,
2.) DegG1✁✄¯
H0
G2
(U0, VI) = DegG2
VI for The Vertex obtained only from VI of
G2 H2,
3.). DegG1✁✄¯
H0
G2
(ui
, VI) = DegG1
ui + DegG2
VI - DegH0 wi for The Vertex obtained
from ui Both of H1 and H2 of VI.
Proof. Let G1 and G2 be any nontrivial graphs, H1 ⊆ G1, H2 ⊆ G2. Let
H0 be labeled The Clone of The labeled graph glued G1✁✄¯
H0
G2 and I be a natural
number. The vertices (ui
, V0), (U0, VI) and (ui
, VI) are The vertices which are obtained
from only ui of G1 H1, only VI of G2 H2 and Both ui of H1 and VI of
H2, respectively. . Since ui
in G1 H1 Becomes (ui
, V0) and VI
in G2 H2 Becomes
(U0, VI) Without The combining of Two Graphs labeled Original, The Degree is Still
The Same. So DegG1✁✄¯
H0
G2
(ui
, V0) = DegG1
ui and DegG1✁✄¯
H0
G2
(U0, VI) = DegG2
VI
. Finally,
we EDGE See that each of H0
contributes Twice in The Degree of (ui
, VI). Thus,
DegG1✁✄¯
H0
G2
(ui
, VI) = DegG1
ui + DegG2
VI - DegH0 wi
. ¥

ผลลัพธ์ (อังกฤษ) 3:

To apply theorem 1.1 in our theorem we must, seek the degree of each vertex
of the glued graph. There are relationships. Between the degree of the vertex of
the, glued graph the two original graphs and the clone that can be stated in the
following. Lemma.
Lemma 2.1. Let G1 and G2 be any nontrivial graphs H1 G1 H2, ⊆, ⊆ G2. Let H0
be the labeled clone of the labeled glued. Graph G1 ✁ ✄ H0 ¯

.G2 and I be a natural number.
Then the degree of each vertex in the labeled glued graph is as follows:
1.) degG1 ✁ ✄ ¯


H0 G2 (UI
,. V0) = degG1
UI for the vertex obtained from only UI of
G1. H1
2,.) degG1 ✁ ✄ ¯


, H0 G2 (U0 VI) = degG2
VI for the vertex obtained. From only VI of
G2. H2
3,.) degG1 ✁ ✄ ¯


H0 G2 (UI
, VI) = degG1

UI degG2 VI − degH0 WI for the vertex obtained
.From both UI of H1 and VI of H2.
Proof. Let G1 and G2 be any, nontrivial graphs H1 ⊆ G1 H2 ⊆, G2. Let
H0 be the labeled. Clone of the labeled glued graph G1 ✁ ✄ ¯

G2 H0 and I be a natural
number. The vertices (UI
, V0), (U0 VI), and (UI
, VI are.) The vertices which are obtained
from only UI of G1. H1 only VI, of G2. H2 and both UI of H1 and VI of
H2 respectively. Since,, UI
in G1. H1 becomes (UI
,V0) and VI
in G2. H2 becomes
(U0 VI), without combining of the two labeled, original graphs the degree is still
the, same. So degG1 ✁ ✄ ¯


H0 G2 (UI
, V0) = degG1
UI and degG1 ✁ ✄ ¯


, H0 G2 (U0 VI) = degG2

.
Finally VI, we see that each edge of H0
contributes. Twice in the degree of (UI
, VI). Thus
degG1, ✁ ✄ ¯


H0 G2 (UI
, VI) = degG1

UI degG2 VI − degH0 wi
.
. Yen