3.1 Triangle-**free** **and** claw-**free** casesWe begin this subsection by giving a simple, but useful, characterization **of**triangle-**free** **2K** 2 -**free** **graphs**.We say that a graph G is **of** C ∗ 5-type if its vertex set can be partitionedinto five disjoint non-empty sets A 1 , A 2 , . . . , A 5 such that each A i is an independentset, A i ∪ A i+1 induces a complete bipartite graph in G for all i(taken modulo 5), **and** the union **of** any other pair **of** sets A i ∪ A j induces anindependent set.Lemma 2. Let G = (V, E) be a connected triangle-**free** **2K** 2 -**free** graph. Theneither G is bipartite or G is **of** C ∗ 5-type.Pro**of**. Suppose G is not bipartite. Then G contains an odd cycle. SinceG is triangle-**free** **and** **2K** 2 -**free**, this implies G contains an induced 5-cycleC = x 1 x 2 x 3 x 4 x 5 x 1 , **and** every vertex **of** G has at most two neighbours inC. If x ∈ V \ V (C) has one neighbour in C, say x 1 , then xx 1 **and** x 3 x 4induce a copy **of** **2K** 2 ; if x has no neighbours in C, then by connectivity xhas at least one neighbour y in G. Since y has at most two neighbours inC, there is some edge x i x i+1 in C such that y is not adjacent to x i **and** x i+1 .But then x i x i+1 **and** xy induce a copy **of** **2K** 2 . Thus every vertex **of** G hasexactly two neighbours in C. Since G is triangle-**free**, every vertex v **of** G hasneighbours x i−1 **and** x i+1 in C for some i (taken modulo 5). Let A i denotethe set **of** vertices adjacent to x i−1 **and** x i+1 so that the A i ’s form a partition**of** the vertices. By triangle-**free**ness, each A i ∪ A i+2 is an independent set,**and** by **2K** 2 -**free**ness, each A i ∪ A i+1 induces a complete bipartite graph in G(if y i ∈ A i **and** y i+1 ∈ A i+1 are non-adjacent, then x i−1 y i **and** y i+1 x i+2 inducea copy **of** **2K** 2 ). Thus G is **of** C ∗ 5-type.Lemma 2 implies the following corollary. We write α(G) for the size **of**the largest independent set **of** G.Corollary 3.Let G be a connected **2K** 2 -**free** graph on n vertices.(i) If G is triangle-**free**, then α(G) ≥ 2n/5.(ii) If G is non-complete **and** 4-tough, then G contains at least two vertexdisjointtriangles.Pro**of**. Statement (i) follows immediately from Lemma 2.6

Suppose, contrary to (ii), that G = (V, E) is a non-complete 4-tough**2K** 2 -**free** graph with no pair **of** vertex-disjoint triangles. Thus we can makeG triangle-**free** by removing at most 3 vertices, **and** so, by (i) **and** usingProposition 1, G has an independent set I with |I| ≥ 2(n − 3)/5 ≥ 2. ThusV \ I is a vertex cut with |V \ I| ≤ 3(n−3)5+ 3. Since G is 4-tough, we have4 ≤|V \ I|ω(G − (V \ I))≤3(n − 3)/5 + 32(n − 3)/5≤ 3 2 + 3 2 = 3,a contradiction.Now we prove that Chvátal’s Conjecture is true for triangle-**free** **2K** 2 -**free****graphs**; moreover, we give the extremal lower bound on **toughness** that guarantees**hamiltonicity**. While the pro**of** is conceptually simple, it is technicallyawkward, **and** this makes it longer than one might expect.Theorem 4. Let G = (V, E) be a triangle-**free** **2K** 2 -**free** graph on at leastthree vertices. Then G is hamiltonian if **and** only if G is 1-tough.Pro**of**. Since 1-**toughness** is clearly a necessary condition for **hamiltonicity**, itsuffices to prove that G is hamiltonian if G is 1-tough. For this, let G be a 1-tough triangle-**free** **2K** 2 -**free** graph on n ≥ 3 vertices. Then G is 2-connected**and** by Lemma 2, G is either bipartite or **of** C ∗ 5-type.First suppose G is bipartite. Since G is 1-tough, the bipartition must beinto two equal-sized sets X = {x 1 , . . . , x k } **and** Y = {y 1 , . . . , y k }. Since G is**2K** 2 -**free**, the neighbourhoods **of** the vertices in X are nested: we can orderthem so that N(x i ) ⊆ N(x j ) for all i ≤ j. Since G is 1-tough, |N(x i )| > i fori = 1, . . . , k − 1 **and** N(x k ) = Y . Thus we can order the vertices **of** Y suchthat N(x i ) ⊇ {y 1 , . . . , y i+1 } for i = 1, . . . , k − 1. This immediately gives usthe Hamilton cycle y 1 x 1 y 2 x 2 · · · y k x k y 1 .Next suppose G is **of** C ∗ 5-type with sets A i **of** cardinality a i as in Lemma 2,where indices are taken modulo 5. We choose an indexing **of** the sets A i insuch a way that a 1 − a 5 = max i |a i+1 − a i |. Note that G is 1-tough if **and**only ifa i−2 − a i−1 + a i − a i+1 + a i+2 ≥ 0 (1)a j−1 − a j + a j+1 ≥ 1 (2)for all i, j ∈ {1, . . . , 5}, because any tough set S **of** G is **of** the form S =A i−2 ∪ A i ∪ A i+2 for some i or **of** the form S = A j−1 ∪ A j+1 for some j.7