The complete short proof of the Berge conjecture

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Definition 1.7 (The berge index and a representative). If G ∈ Ω, then the berge index of G is denoted by β(G) where β(G) = min F ∈B(G) ω(F ) and where B(G) = {F ; G ∈ parent(F ) and F is berge} [B(G) is the set of graphs F such that G is a parent of F and F is berge]; and a representative of G is any graph S ∈ B(G) such that ω(S) = β(G). If G ∈ Ω, then the berge index of G is denoted by β(G) where β(G) = min P ∈parent(G) β(P ) and where parent(G) is the set of all parents of G [use Definition 1.3]; and a representative of G is any graph S such that ω(S) = β(G) and S is an element of any set B(P ) such that P ∈ parent(G) and β(P ) = β(G) [ notice that B(P ) = {F ; P ∈ parent(F ) and F is berge}] [we will prove that the berge index β(G) exists and is well defined for every graph G and we will also prove that for every graph G, there exists at least one representative of G].
Using Definitions 1.7, let us Remark.
Remark 1.7 . Let G be a graph. Then β(G) exists, is well defined and there exists S such that S is a representative of G [such a S is berge and we have χ(S) = χ(G)].
Proof. (1.1) Using (1.1) and the meaning of ( β(G), B(G) ), then we immediately deduce that β(G) exists, is well defined, S is a representative of G and S is berge and χ(S) = χ(G).
(1.2) (ii) If G ∈ Ω, then β(G) exists, is well defined and there exists a graph S such that S is a representative of G [such a S is berge and we have χ(S) = χ(G)]. (Indeed let P be a parent of G [such a P exists by using Assertion 1.4]; since in particular P ∈ Ω [because P is a parent of G] , then using (i), we immediately deduce that β(P ) exists and therefore min P ∈parent(G) β(P ) also exists. Remark 1.7 . Let (K, G, B, P ), where K is a complete graph, G ∈ Ω, B is berge and P ∈ parent(B ). We have the following elementary properties.
(1.7 .2). Indeed let B(K) = {F ; K ∈ parent(F ) and F is berge}, since K is a complete graph, clearly K is berge and B(K) = {K}; observe that K ∈ Ω and so β(K) = min ω(F ) [use the definition of the parameter β and observe that G ∈ Ω]; observe that G is berge [use Assertion 1.6], so G ∈ B(G) and the previous equality implies that ω(G) ≥ β(G).Property (1.7 .3) follows.
(1) ⇒ (2). Indeed let F be graph and let S be a representative of F , in particular S is berge [use Remark 1.7 ] and clearly χ(S) = ω(S); now observing that ω(S) = β(F ) [since S is a representative of F ], then the previous two equalities imply that χ(S) = β(F ). Observe that χ(S) = χ(F ) [by observing that S is a representative of F and by using Remark 1.7 ] and the last two equalities immediately become χ(F ) = β(F ).
(3) For every G ∈ Ω such that 0 ≤ χ(G) ≤ 2, we have ω(G) = β(G). 2 www.ijc.or.id The complete short proof of the Berge conjecture | Ikorong Annouk 2. Uniform graphs, relative subgraphs, the berge caliber and some consequences In this section, we use the original reformulation of the Berge problem given by Theorem 1.8 to introduce uniform graphs and relative subgraphs [uniform graphs and relative subgraphs are crucial for the proof of the result which implies the Berge conjecture], and we give some elementary properties of these graphs. In this section, we also define another graph parameter denoted by b [the graph parameter b is called the berge caliber, and is related to the berge index defined in Section 1], and using the graph parameter b, we prove simple propositions that we will use in Section 4 to give the short proof of the Berge problem [in particular, we prove a trivial proposition which is equivalent to the Berge problem] Remark 2.1. Let K be a complete graph; then β(K) = ω(K) = χ(K). Proof. Immediate and is an obvious consequence of property (1.7 .2) of Remark 1.7 . 2 From Theorem 1.8, we are going to define a new class of graphs in Ω [called uniform graphs]; we will also define relative subgraphs, and we will present some properties related to these graphs. These properties are elementary and curiously, are crucial for the result which immediately implies the Berge problem and the Berge conjecture. Before, let us define.
Definition 2.2 [optimal coloration and (G)]. An optimal coloration of a graph G is a partition is the chromatic number of G]; (G) denotes the set of all optimal colorations of G; so, Ξ(G) ∈ (G) means Ξ(G) is an optimal coloration of G. Using the denotation of (G) [ use Definition 2.2], then the following Assertion is immediate.
Definition 2.5 (Uniform graph) Let G ∈ Ω and let Ξ(G) be the canonical coloration of G [observe that Ξ(G) exists, via Assertion 2.4]; we say that G is uniform, if for every Y ∈ Ξ(G), we have card(Y ) = α(G), where card(Y ) is the cardinality of Y and where α(G) is the stability number of G.
Definition 2.5 gets sense, since G ∈ Ω and so Ξ(G) is canonical [via Assertion 2.4]. Using the definition of a uniform graph [use Definition 2.5], then the following Assertion is immediate.
Assertion 2.6. Let G ∈ Ω and let Ξ(G) be the canonical coloration of G [observe that Ξ(G) exists, via Assertion 2.4]. We have the following trivial properties. (2.6.0). If 0 ≤ ω(G) ≤ 1, then G is uniform. (2.6.1). If 0 ≤ α(G) ≤ 1, then G is uniform. (2.6.2). If G is a complete graph, then G is uniform. (2.6.3). If α(G) ≥ 2 and if for every Y ∈ Ξ(G) we have card(Y ) = α(G), then G is uniform and is not a complete graph. Proof. Properties (2.6.0) and (2.6.1) are trivial [it suffices to observe that G ∈ Ω and to use Definition 2.5]; property (2.6.2) is an immediate consequence of property (2.6.1); and property (2.6.3) is trivial (indeed, observe [by the hypotheses] that G ∈ Ω and use Definition 2.5). 2 Uniform graphs have nice properties when we study isomorphism of graphs. Recall 2.7. Recall that two graphs are isomorphic if there exists a one to one correspondence between their vertex set that preserves adjacency.
Assertion 2.8. Let G ∈ Ω; then there exists a uniform graph U which is isomorphic to a parent of G [use Definition 1.3 for the meaning of parent]. Proof. If 0 ≤ ω(G) ≤ 1, clearly G is uniform [use property (2.6.0) of Assertion 2.6]; now put U = G, clearly U is a uniform graph which is a parent of G. Now, if ω(G) ≥ 2, let Ξ(G) be the canonical coloration of G [observe that Ξ(G) exists, by remarking that G ∈ Ω and by using Assertion 2.4]; since it is immediate that χ(G) = ω(G), clearly Ξ(G) is of the form Ξ(G) = {Y 1 , ..., Y ω(G) }. Now let Q be a graph defined as follows: (i) Ξ(Q) = {Z 1 , ..., Z ω(G) } is a partition of V (Q) into ω(G) stable sets such that, x ∈ Z j ∈ Ξ(Q), y ∈ Z k ∈ Ξ(Q) and j = k, ⇒ x and y are adjacent in Q; (ii) For every j = 1, 2, ..., ω(G) and for every , and visibly, G is isomorphic to a subgraph of Q; observe that Q is isomorphic to a true pal of G and Q ∈ Ω [since G is isomorphic to a subgraph of Q and χ(Q) = ω(Q) = ω(G) = χ(G) and Q ∈ Ω] and Q is uniform. Using the previous and the definition of a parent, then we immediately deduce that Q is a uniform graph which is isomorphic to a parent of G. Now put Q = U ; clearly U is a uniform graph which is isomorphic to a parent of G. Assertion 2.8 follows. 2 Now the following assertion is only an immediate consequence of Assertion 2.8. Assertion 2.9. Let H be a graph [H is not necessarily in Ω]; then there exists a uniform graph U which is isomorphic to a parent of H. Proof. Let P be a parent of H [such a P exists via Assertion 1.4] and let U be a uniform graph such that U is isomorphic to a parent of P [such a U exists, by observing that P ∈ Ω and by using Assertion 2.8]; clearly U is a uniform graph and is isomorphic to a parent of H [since U is a uniform graph which is isomorphic to a parent of P and P is a parent of H] . 2 Now we define relative subgraphs.
Definition 2.10 (relative subgraph). Let G and F be uniform. Now let Ξ(G) be the canonical coloration of G and let Ξ(F ) be the canonical coloration of F [observe that the couple (Ξ(G), Ξ(F )) exists, by remarking that G ∈ Ω and F ∈ Ω, and by using Assertion 2.4]. We say that F is a Assertion 2.11. Let (P, U ) be a couple of uniform graphs such that ω(P ) ≥ 1 and ω(U ) ≥ 1. Now let Ξ(P ) be the canonical coloration of P and let Ξ(U ) be the canonical coloration of U [observe that the couple (Ξ(P ), Ξ(U )) exists, by remarking that P ∈ Ω and U ∈ Ω, and by using Assertion 2.4]. Then we have the following trivial properties. (2.11.0). If U is a relative subgraph of P , then α(U ) = α(P ) and ω(U ) ≤ ω(P ). (2.11.1). If U is a relative subgraph of P and if ω(U ) = ω(P ) , then U = P . (2.11.2). If U is a relative subgraph of P and if ω(U ) < ω(P ), then there exists Y ∈ Ξ(P ) such that U is a relative subgraph of P \ Y . (2.11.3). If α(U ) = α(P ) and ω(U ) = ω(P ), then U and P are isomorphic. (2.11.4). If ω(P ) ≥ 2, then, for every Y ∈ Ξ(P ), P \ Y is a relative subgraph of P and P \ Y is uniform and ω(P \ Y ) = ω(P ) − 1 and α(P \ Y ) = α(P ).
Proof. Properties (2.11.0) and (2.11.1) and (2.11.2) are immediate [it suffices to use the definition of a uniform graph and the definition of a relative subgraph]. Properties (2.11.3) and (2.11.4) are trivial consequences of the definition of uniform graphs and relative subgraphs. 2 We will see in Section 4 that uniform graphs and relative subgraphs play a major role in the proof of Theorem which immediately implies the Berge problem. Now we introduce again definitions that are not standard; in particular, we introduce a graph parameter denoted by b and called the berge caliber [the berge caliber b is related to the berge index β introduced in Definitions 1.7 (Section 1)], and using the parameter b on uniform graphs, we prove an elementary proposition which will help us in Section 4 to prove the Berge problem and the Berge conjecture. Before, let us define. The following two assertions are obvious consequences of Remark 2.1 and Definition 2.12. Assertion 2.13. Let K be a complete graph; then K is bergerian. Proof. Immediate, and is a consequence of Remark 2.1 and Definition 2.12 and property (2.6.2) of Assertion 2.6.2 Assertion 2.14. The set of all complete graphs is an obvious example of bergerian graphs. Proof. Immediate, and is a trivial consequence of Assertion 2.13.2 Definitions 2.15 (bergerian subgraph and maximal bergerian subgraph). Let G be uniform. We say that a graph F is a bergerian subgraph of G, if F is bergerian and is a relative subgraph of G [use Definition 2.10 for the meaning of a relative subgraph and Definition 2.12 for the meaning of bergerian ]. We say that F is a maximal bergerian subgraph of G [we recall that G is uniform], if F is a bergerian subgraph of G and ω(F ) is maximum for this property [it is immediate that such a F exists and is well defined]. Now we define the berge caliber.
The following remark clearly shows that for every uniform graph G, b(G) exists and is well defined.
Remark 2.17.For every uniform graph G, the berge caliber b(G) exists and is well defined. Proof. Let G be uniform and let F be a maximal bergerian subgraph of G [use Definitions 2.15]; observing [by definition of a maximal bergerian subgraph of G] that F is a bergerian subgraph of G and ω(F ) is maximum for this property, clearly ω(F ) is unique and therefore b(G) is also unique, since b(G) = ω(F ). So b(G) exists and is well defined. 2 It is immediate that the berge caliber [i.e. the graph parameter b] is defined for uniform graphs and only for uniform graphs. We will see in Section 4 that the berge caliber plays a crucial role in the proof of Theorem which immediately implies the Berge problem. Now, using the definition of a uniform graph, the definition of a relative subgraph and the definition of the berge caliber, then the following assertion is immediate.  The previous definitions and simple properties made, now the following trivial proposition is crucial for the proof of the Berge problem and the Berge conjecture. Now let B be berge and let P be uniform such that P is isomorphic to a parent of B [such a P clearly exists, via Assertion 2.9], clearly [by observing that in particular P is isomorphic to a parent of B and by using Remark 2.0]. It trivial that Since in particular P is isomorphic to a parent of B, clearly P is isomorphic to a true pal of B and so Clearly ω(P ) = χ(P ) [since P ∈ Ω], and using the previous equality, then it becomes trivial to deduce that equality (2.4) clearly says that ω(P ) = χ(B).

(2.5)
Recalling that P is uniform and using (2.1), then it becomes trivial to deduce that Now using (2.2) and (2.3) and (2.5) and (2.6), then it becomes trivial to deduce that The complete short proof of the Berge conjecture | Ikorong Annouk (2.7) immediately implies that , and the previous equality clearly says that the Berge problem is true for B; using the previous and observing that the berge graph B was arbitrary chosen, then it becomes trivial to deduce that every berge graph B satisfies ω(B ) = χ(B ); so the Berge problem is true and therefore (i) ⇒ (ii).
, then, using Theorem 1.8, we immediately deduce that Now let U be uniform; observing that U ∈ Ω and using (2.9), then we immediately deduce that for every uniform graph U, we have ω(U ) = β(U ). (2.10) Now using (2.10) and property (2.18.1) of Proposition (2.18), then it becomes trivial to deduce that for every uniform graph U , we have ω In Section 4, Proposition 2.19 coupled with trivial complex calculus and elementary computation we help us to give the short proof of the Berge problem and the Berge conjecture.
3. Isomorphism of uniform graphs, properties linked to trivial complex calculus on uniform graphs, elementary computation, and the reduction of the Berge problem into a simple equation of three unknowns In this section, we prove elementary properties linked to trivial complex calculus on uniform graphs and elementary computation; and we reduce the Berge problem into a simple equation of three unknowns. We will use them in Section 4, where we will give the short complete simple proof of the Berge problem and the Berge conjecture [In this Section, we let one Proposition unproved and we will prove this Proposition in Section 3' (Epilogue)]. In this section, the definition of a uniform graph [use Definition 2.5 ], the definition of a relative subgraph [use Definition 2.10 ], the definition of the berge caliber b(G) of a uniform graph G [use Definition 2.16 ], are crucial. Before, let us remark that uniform graphs and relative subgraphs have interesting properties when we study isomorphism of graphs. Indeed, using the definition of uniform graphs and the definition of relative subgraphs and the definition of isomorphism [use Remark 2.7], then the following Remark becomes immediate: Remark 3.0. Let U and U be uniform graphs such that ω(U ) ≥ 2 and ω(U ) ≥ 1. Now let Ξ(U ) be the canonical coloration of U [observe that Ξ(U ) exists, by remarking that U ∈ Ω (since U is uniform) and by using Assertion 2.4], and let (Y, Then we have the following four simple properties. (i) U and U are isomporphic, if and only if, ω(U ) = ω(U ) and α(U ) = α(U ).
(ii) U is isomorphic to a relative subgraph of U if and only if ω(U ) ≥ ω(U ) and α(U ) = α(U ).
Proof. Property (i) is immediate [indeed, observe that U and U are uniform graphs and use the definition of uniform graphs]. Property (ii) is trivial [indeed, observe that U and U are uniform graphs and use the definition of uniform graphs and the definition of relative subgraphs]. Property Proposition 3.1. Let U be uniform such that ω(U ) ≥ 2 and let Ξ(U ) be the canonical coloration of U [observe that Ξ(U ) exists, by remarking that U ∈ Ω (since U is uniform) and by using Assertion It is trivial that (3.1) clearly says that Recalling that F is a maximal bergerian subgraph of U [ and so ω(F ) = b(U )] and using equality (3.2), then it becomes trivial to deduce that Proposition 3.1 will help us to simplify complex calculus on uniform graphs.

Recalls and Definitions 3.2 (Fundamental). (Real numbers, complex numbers, C, L U and tackle).
Recall that R is the set all real numbers and θ is a complex number if θ = x + iy, where x ∈ R, y ∈ R and i 2 = −1; C is the set of all complex numbers. That being said, clearly Now let n be an integer ≥ 1 and let U be uniform such that ω(U ) = n; look at the berge caliber We will see that the definition of tackle introduced above helps to reduce the Berge problem into an elementary equation of three unknowns. Before, let us define: Let n be an integer ≥ 1 and let U be a uniform graph such that ω(U ) = n; look at L U [use Recalls and and where and and It is immediate that for every integer n ≥ 1 and for every uniform graph U such that ω(U ) = n, (φ(U ), ν(U ), (U )) is well defined and gets sense. Now using the notion of tackle (use Recalls and Definition 3.2), then the following Theorem immediately implies the Berge problem.
Theorem B. Let n be an integer ≥ 1 and let (U, Q) be a couple of uniform graphs such that ω(Q) = n + 1 and U is a relative subgraph of Q with ω(U ) = n; Then at least one of the following three properties is satisfied. We will simply prove Theorem B in Section 4. But before, let us remark.
Remark 3.4 (fundamental) Let (n, U ) where n is an integer ≥ 1 and U is a uniform graph such that ω(U ) = n; consider L U (use Recalls and Definitions 3.2) and look at ( Proof. Indeed, observing (via the hypotheses) that L U = 2n + 2 and using the preceding equality, we easily deduce that (use (3.6) of Definitions 3.3 and the third equality of (3.11) (observe [via the third equality of (3.11)] that (L U − 1) 2 = L 2 U − 2L U + 1); and the first equality of (3.11) (observe [via the first equality of (3.11)] that (L U −2n) 2 = 1)), and (use (3.7) of Definitions 3.3 and the third equality of (3.11) (observe [via the third equality of (use (3.8) of Definitions 3.3 and the third equality of (3.11) (observe [via the third equality of (3.11)] that (L U −1) 2 = L 2 U −2L U +1)). That being said let (ν(U ), (U )) explicited in Definitions www.ijc.or.id The complete short proof of the Berge conjecture | Ikorong Annouk (use (3.9) of Definitions 3.3 and the second equality of (3.11)), and clearly (use (3.10) of Definitions 3.3 and the last two equalities of (3.11)). That being so, let (x, y, k) such that .15) and (3.16) and the first two equalities of (3.17), we easily check (by elementary computation and the fact that That being so, look again at (φ(U ), ν(U ), (U ), x, y, k ) and let ( φ(U ), x, y, k ); using the three equalities of (3.17), it becomes very easy to check (by elementary computation and the fact that  (3.18) and (3.19) and the notion of tackle introduced in Recalls and Definitions 3.2 [observe that (x, y) ∈ C 2 and k ∈ R; so (x, y, k) ∈ C 2 × R]). Remark 3.4 follows.
Remark 3.5 (fundamental): reduction of the Berge problem into a trivial equation of three unknowns. Let (n, U ) where n is an integer ≥ 1 and U is a uniform graph such that ω(U ) = n; consider L U (use Recalls and Definitions 3.2) and look at ( φ(U ), ν(U ), (U ) ) introduced in Recalls and Definitions 3. Proof. Indeed, observing (via the hypotheses) that L U = 2n and using the preceding equality, we easily deduce that The complete short proof of the Berge conjecture | Ikorong Annouk (use (3.6) of Definitions 3.3 and the third equality of (3.20) (observe [via the third equality of (3.20)] that (L U + 1) 2 = L 2 U + 2L U + 1); and the first equality of (3.20) (observe [via the first equality of (3.20)] that (L U −2n) 2 = 0)), and  That being said, we have this fact.
Otherwise (we reason by reduction to absurd) It is immediate to see that (3.28) says that (use (3.4) for φ(U )). Now let ( φ(U ), ν(U ), (U ), x, y ) explicited above; consider (φ(U ), x, y) and look at (3.29); using elementary computation and elementary divisibility coupled with the fact that i 2 = −1 and k ∈ R, it becomes very easy to check that (3.29) immediately implies that and

So assuming that
there exists k ∈ R such that x + 3iyL −1 gives rise to a serious contradiction; therefore    Observe that Theorem 4.1 is exactly Theorem.B stated after Definitions 3.3 (Section 3). We are going to prove simply Theorem 4.1. But before, let us remark.
Remark 4.1 . Let n be an integer ≥ 2 and let U be uniform such that ω(U ) = n; consider the canonical coloration Ξ(U ) of U [observe that Ξ(U ) exists, by remarking that U ∈ Ω (since U is uniform) and by using Assertion 2.4]. Now let Y ∈ Ξ(U ) and look at the couple [note that U is bergerian and use property (2.18.1) of Assertion 2.18], and notice that U \ Y is unif orm such that ω(U \ Y ) = ω(U ) − 1 = n − 1 and n − 1 ≥ 1 (4.1) [note that U is uniform such that n = ω(U ) ≥ 2, Y ∈ Ξ(U ), and use property (iii) of Remark [use (4.0) and (4.1) and the fact that U is bergerian]. So U \ Y is bergerian [ note that U \ Y is uniform such that ω(U \ Y ) = n − 1 ≥ 1 ( use (4.1) ) and use (4.2) coupled with property (2.18.1) of Assertion 2.18]. Remark 4.1 follows. 2 Remark 4.1 . Let n be an integer ≥ 2 and let Q be uniform such that ω(Q) = n. Now let U be a relative subgraph of Q. If Q is bergerian, then U is also bergerian .
Proof. (We reason by reduction to absurd). Indeed let Q be a counter-example such that card(V (Q)) is minimun; clearly U = Q and ω(Q) ≥ 3 (4.2 ) [otherwise U = Q or ω(Q) = 2, and it becomes immediate to deduce that U is bergerian; a contradiction]. Now using (4 .2 ) and the definition of relative subgraph, then we immediately deduce that there exists Y ∈ Ξ(Q) such that U is a relative subgraph Q \ Y and card( V (Q \ Y ) ) < card(V (Q)).

(4.2 )
Clearly U is bergerian [use (4.2 ) and the minimality of card(V (Q)) and Remark 4.1 ]; a contradiction. Remark 4.1 follows. 2 Remark 4.2. Let n be an integer ≥ 1 and let (U, Q) be a couple of uniform graphs such that ω(Q) = n + 1 and U is a relative subgraph of Q with ω(U ) = n; consider (b(U ), b(Q)) ( where b(U ) is the berge caliber of U and b(Q) is the berge caliber of Q). We have the following trivial properties. Observing that U is a relative subgraph of Q with ω(U ) + 1 = ω(Q), then we immediately deduce that [use (4.4) and (4.5) and Remark 4.1 ]. Observe that [use (4.6) and property (2.18.1) of Assertion 2.18]. So [use (4.3) and (4.7) and the fact that ω(Q) = ω(U ) + 1 = n + 1]. (4.7) clearly says that property (A 2 ) of Theorem 4.1 is satisfied; therefore Theorem 4.1 is also satisfied. Property (4.2.1) follows). Property (4.2.2) is immediate (indeed if b(U ) = n, observing that n ≥ 1 and U is uniform such that ω(U ) = n, then using the previous and the denotation of L U given in Recall and Definition 3.2, we clearly deduce that L U = 2n + 2; now using the previous equality and Remark 3.4, we immediately deduce that (4.8) (4.8) clearly says that property (A 3 ) of Theorem 4.1 is satisfied; therefore Theorem 4.1 is also satisfied. Property (4.2.2) follows). Property (4.2.3.) is also immediate (indeed, observing (by the hypotheses) that n = 1, clearly 2 = n + 1 = ω(Q). (4.9) Since Q is uniform, then, using (4.9) and property (2.18.3) of Assertion 2.18, it becomes very easy to deduce that n + 1 = ω(Q) = b(Q) = β(Q). (4.10) Clearly Since U is uniform, then, using (4.12) and property (2.18.3) of Assertion 2.18, it becomes very easy to deduce that (4.13) Clearly b(U ) = n (4.14) [use (4.13)]. So Theorem 4.1 is satisfied [use (4.14) and property (4. Using Remark 4.3, then the following last definition comes. Definition 4.4 (Fundamental). Suppose that Theorem 4.1 is false and let n be a minimum counter-example to Theorem 4.1 (such a n exists, by using Remark 4.3). We say that (U, Q) is a remarkable couple of uniform graphs, if (U, Q) is a couple of uniform graphs such that ω(Q) = n+1 and U is a relative subgraph of Q with ω(U ) = n [n ≥ 1 and n is a minimum counter-example to Otherwise (we reason by reduction to absurd), let n be a minimum counter-example to Theorem 4.1 ( such a n exists, by using Remark 4.3), and let (U, Q) be a remarkable couple of uniform graphs ( such a couple (U, Q) exists, by using Remark 4.3 and Definition 4.4); fix once and for all the couple (U, Q) ( the couple (U, Q) is fixed once and for all, so (U, Q) does not move anymore), and look at ( φ(U ), ν(U ), (U ) ) introduced in Definitions 3.3. We observe the following. Observation.4.1.i. n + 1 = ω(Q); n = ω(U ) and ( φ(U ), ν(U ), (U ) ) does not tackle (1, 3iL −1 and Ξ(U ) ⊆ Ξ(Q). [Indeed recalling that Y ∈ Ξ(U ) and observing that U is a relative subgraph of Q such that ω(Q) = ω(U ) + 1 (because (U, Q) is a couple of remarkable uniform graphs), then using the preceding and (4.16) via the meaning of a relative subgraph ( use Definition 2.10), it becomes immediate to deduce that Q\Y and U \Y are relative subgraphs of Q and U \Y is a relative subgraph of Q \ Y and ω(Q \ Y ) = ω(Q) − 1 = n and ω(U \ Y ) = ω(U ) − 1 = n − 1 and n − 1 ≥ 2 and Ξ(U ) ⊆ Ξ(Q)]. Observation.4.1.iv follows.
Observation 4.1.v. Let n [we recall n is a minumun counter-example to Theorem 4.1] and look at the couple (U, Q) [ we recall that (U, Q) is a couple of remarkable uniform graphs and is fixed once and for all ]. Now let Ξ(Q) be the canonical coloration of Q [observe that Ξ(Q) exists, by remarking that Q ∈ Ω (since Q is uniform) and by using Assertion 2.4]. Then there exists Z ∈ Ξ(Q) such that Z ∈ Ξ(U ) and U is isomorphic to Q \ Z (note that G \ Y is the induced subgraph of G by V (G) \ Y and Ξ(U ) is the canonical coloration of U (observe that Ξ(U ) exists, by remarking that U ∈ Ω (since U is uniform) and by using Assertion 2.4).
Indeed observing that U is a relative subgraph of Q such that ω(Q) = ω(U ) + 1 ( because (U, Q) is a couple of remarkable uniform graphs), then using the preceding and the meaning of a relative subgraph [ use Definition 2.10], we immediately deduce that there exists Z ∈ Ξ(Q) such that Z ∈ Ξ(U ). [use (4.19) and the the meaning of a relative subgraph (use Definition 2.10) and the last two equalities of (4.18)]. Since in particular (U, Q) is a couple of uniform graphs such that U is a relative subgraph of Q with ω(Q) = ω(U ) + 1, then using the previous and the couple ((4.19), (4.20)), it becomes immediate to deduce that is a couple of uniform graphs and Observation 4.1.vi. Let n [we recall n is a minumun counter-example to Theorem 4.1] and look at the couple (U, Q) [we recall that (U, Q) is a couple of remarkable uniform graphs and is fixed once and for all]. Now let Ξ(U ) be the canonical coloration of U [observe that Ξ(U ) exists, by remarking that U ∈ Ω (since U is uniform) and by using Assertion 2.4]; look at Y ∈ Ξ(U ) and let Firstly, we prove that U \ Y is uniform and ω(U \ Y ) = ω(U ) − 1 = n − 1 and ω(Q \ Y ) = ω(Q) − 1 = n. Fact.0: U \Y is uniform and ω(U \Y ) = ω(U )−1 = n−1 and ω(Q\Y ) = ω(Q)−1 = n. Indeed, using Observation.4.1.iv, then it becomes immediate to deduce that U \ Y is a relative sub- Secondly, we prove that b(U ) ≤ n − 1. [since (U, Q) is a couple of remarkable uniform graphs], then, using (4.29) and (4.30) and Remark 4.1 , we immediately deduce that U is bergerian.  www.ijc.or.id The complete short proof of the Berge conjecture | Ikorong Annouk Observation.4.1.vii follows.
Observation 4.1.viii. Let n [ we recall n is a minumun counter-example to Theorem 4.1] and look at the couple (U, Q) [ we recall that (U, Q) is a couple of remarkable uniform graphs and is fixed once and for all ]. Let Ξ(U ) be the canonical coloration of U and let Y ∈ Ξ(U ); look at Indeed, look at n [recall n is a minimum counter-example to Theorem 4.1 ], and via n, consider n−1 (this consideration gets sense, since n ≥ 3 [use Observation.4.1.iii], and therefore n−1 ≥ 2).
Then by the minimality of n, n − 1 is not a counter-example to Theorem 4.1. [use (4.43)], then using (4.46) and (4.46 ) and the fact that U is uniform, it becomes immediate to deduce that U is a complete graph. Otherwise [use Observation.4.1.vi ], then using (4.49), it becomes immediate to deduce that (4.48) clearly says that Having made the previous three Claims, then we immediately deduce that [use (4.56) and (4.53)]. Observation.4.1.viii follows.
Observation 4.1.ix. Let the couple (U, Q) [we recall that (U, Q) is a couple of remarkable uniform graphs and is fixed once and for all ] and look at ( φ(U ), ν(U ), (U ) ) introduced in Definitions 3.3.
. Indeed using Observation.4.1.vii and the definition of tackle (use Recalls and Definitions 3.2), then we immediately deduce that there exists ( x, y, k ) ∈ C 2 × R such that That being so, using Observation.4.1.viii and the definition of tackle (use Recalls and Definitions 3.2), then we immediately deduce that there exists (x , y , k ) ∈ C 2 × R such that (4.60) Now using equality of (4 .59 ), then we immediately deduce that equality of (4.57) clearly says that It is trivial to see that equality (4.61) clearly says that Using the two equalities of (4.60), then we immediately deduce that the two equalities of (4.58) clearly say that It is trivial to see that (4.63) clearly says that Now observing that ( x, y, k ) ∈ C 2 × R (use (4.57)) and since ( x , y , k ) ∈ C 2 × R (use (4.59)), then it becomes trivial to deduce that (4.62) and (4.64) clearly say that there exists (x , y , k ) ∈ C 2 × R such that x = x − x and y = y − y and k = k − k (4.65)   (4.75) Now observing that U is uniform and n > 2 [use Observation.4.5.1], then using the preceding and (4.75), it becomes trivial to see that (n, U ) satisfies all the hypotheses of Remark 3.5, therefore (n, U ) satisfies the conclusion of Remark 3.5; consequently ( φ(U ), ν(U ), (U ) ) does not tackle www.ijc.or.id The complete short proof of the Berge conjecture | Ikorong Annouk Observation.4.5.2 follows.
(⇐) Otherwise [we reason by reduction to absurd]. Let F be a counter-example such thatcard(V (F )) is minimum. Then, we have the following three simple claims. Indeed, recalling that F is berge, clearly F is also berge [ since F is an induced subgraph of F ]. Since card(V (F )) = card(V (F )) − 1, clearly F is berge and is such that card(V (F )) = card(V (F )) − 1; then, by the minimality of card(V (F )), F is not a counter-example. So, F is perfect. Claim.4.7.1 follows.
Otherwise, clearly These three simple claims made, consider F ; recalling that F is berge, then Theorem 4.6 immediately implies that χ(F ) = ω(F ), and this contradicts claim 4.7.2. Theorem 4.7 follows. 2

3'. Epilogue
Our article clearly shows that elementary complex calculus coupled with elementary arithmetic calculus and trivial computation help to give a simple analytic proof of the Berge problem and the Berge conjecture. Now we end this article by proving the only Proposition that we let unproved in Section 3. Proposition 3.6. Let (n, U ) where n is an integer ≥ 2 and U is a uniform graph such that ω(U ) = n; consider L U (use Recalls and Definitions 3.2) and look at ( φ(U ), ν(U ), (U ) ) introduced in Definitions 3.3. Now let Ξ(U ) be the canonical coloration of U (use Definition 2.3), ) (this consideration gets sense, since U is uniform such that ω(U ) = n with n ≥ 2 and therefore U \ Y is uniform such that ω(U \ Y ) = n − 1 [use property (2.11.4) of Assertion 2.11], with n − 1 ≥ 1).
To prove Proposition 3.6, we need four elementary remarks. (this consideration gets sense, since U is uniform such that ω(U ) = n with n ≥ 2 and therefore U \ Y is uniform such that ω(U \ Y ) = n − 1 [use property (2.11.4) of Assertion 2.11], with n − 1 ≥ 1). If L U ≤ 2n, then we have the following four properties. Proof.