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I need help to understand this proof since the wording is really difficult, I only need some one to explain me from 0.3 in some

I need help to understand this proof since the wording is really difficult, I only need some one to explain me from 0.3 in some easy words

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(0.2) 2 pry-I : 2deg(v) 1- 1X2] + 2, consider the subgraph H made by all edges of colours 1 and 2. Each component of H is a path or circuit. At least one component of H contains more vertices in X1 than in X2. This component is a path P starting in X1 and not ending in X 2 Exchanging colours 1 and 2 on P reduces IX 1[2 + IX2|2, contradicting our minimality assumption. This , proves (3). This implies that there exists an 1' with 1X l = 1, since otherwise by (2) and (3) each [X l is 0 or 2, while their sum is odd, a contradiction So we can assume |Xk| = 1, say X], z: {11}. Let G' be the graph obtained from G by deleting edge vu and deleting all edges of colour is. So G" v is (k 1)-edge-coloured. Moreover, in G' , vertex v and all its neighbours have degree at most k: - 1, and at most one neighbour has degree k ' 1. So' by the induction hypothesis, 6" is (k 1)edgecolourable. Restoring colour Is, am? giving edge vu colour 19, gives a k-edge-colouring of G. . I " homepages.cwi.nl 7 Wmmewtnwex/l'; , i1 4. Theorem (Vizing's theorem for simple graphs). A(G) S x' (G) g A(G) + 1 for any simple graph G. Proof. The inequality A(G) g x' (G) being trivial, we show x' (G) g A(G) + 1. To prove this inductively, it suffices to Show for any simple graph G: (0.1) Let 1) be a vertex such that v and all its neighbours have degree at most k, while at mostone neighbour has degree precisely h. Then if G - 'v is kedge-colourable, also G is k-edgecolourable. Wesprove (1) by induction on k. We can assume that each neighbour u of v has degree k 1, except for one of degree k, since otherwise we can add a new vertex w and an edge uw without violating the conditions in (1). We can do this till all neighbours of 1) have degree k 1, except for one having degree 14:. ' ' Consider any k-edgecolouring of G - '1). For 2' = 1, . . . ,k, let X1; be the .I set of neighbours of 1} that are missed by colour 2'. So all but one neighbour div is in precisely two of the Xi, and one neighbour is in precisely one Xi. ' Hence ' Hi\": \"1 k (0.2) H 'i ' Zile-l = 2deg(v) 1

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