, In this subsection we prove that, at the difference from other hyperbolicity parameters

, Note that if we are given a BFS-tree T rooted at w, we can easily check whether ? w,T (G) ? 1, and thus deciding whether ? ? (G) ? 1 is in NP

, Up to preprocessing the formula, we can suppose that ? satisfies the following properties (otherwise, ? can be reduced to a formula satisfying these conditions): ? no clause c j can be reduced to a singleton, vol.1

, ? every literal x i , x i is contained in at least one clause

, ? no clause c j can be strictly contained in another clause c k

, ? every clause c j is disjoint from some other clause c k (otherwise, a trivial satisfiability assignment for ? is to set true every literal in c j )

, ? if two clauses c j , c k are disjoint, then there exists another clause c p that intersects c j in exactly one literal, and similarly, that also intersects c k in exactly one literal (otherwise, we add the two new clauses x ? y and x ? y, with x, y being fresh new variables; then, we replace every clause c j by the two new clauses c j ? x ? y and c j ? x ? y)

X. Let and .. .. , For simplicity, in what follows, we often denote x i , x i by 2i?1 , 2i . Let C := {c 1 , . . . , c m } be the clause-set of ?. Finally, let w and V = {v 1 , v 2 , . . . , v 2n } be additional vertices. We construct a graph G ? with V (G ? ) = {w} ? V ? X ? C and where E(G ? ) is defined as follows: ? N (w) = V and V is a clique, ? for every i, i, = i

, ? for every i, i , i and i are adjacent if and only if i = i

, ? for every i, j, v i and c j are not adjacent

, ? for every i, j, i and c j are adjacent if and only if i ? c j

, By construction, d(c j , c k ) = 2, and the parent node of c k contained in c j ? c p . We have (c k | i ) c j = 2 ? d(c k , i )/2 ? {1, 3/2}. In particular, (c k | i ) c j = 1. As a result

, For every BFS-tree T rooted at i ? X, we have ? i

, Since there is a perfect matching between X and V = N (w), the parent node of w must be v i . We claim that the parent node of v i , for every i = i, must be also v i . Indeed, otherwise this should be i . However, (w|v i ) i = (2 + 2 ? 1)/2 = 3/2. In particular, (w|v i ) i = 1, and so, ? i ,T (G ? ) ? 1 implies v i and i should be adjacent, that is a contradiction. So, the claim is proved. Then, let c j ? C be nonadjacent to i . We have d(c j , i ) = 2, and the parent node p j of c j, vol.1

, For every BFS-tree T rooted at v i ? V , we have ? v i

, In particular, the parent of i must be v i . Furthermore, there exists c j ? C such that d(v i , c j ) = 2. In particular, i must be the parent of c j . However, ( i |c j ) v i = 2 ? d( i , c j )/2 ? {1, 3/2}. In particular, ( i |c j ) v i = 1. So, ? v i ,T (G ? ) ? d(v i , i ) = 2. From now on, There exists i = i such that i , i are nonadjacent

, If s ? V and t is arbitrary

, Since w is a simplicial vertex and s ? N (w), we have d(w, t) ? 1 ? d(s, t) ? d(w, t), and consequently

, We have (s|t) w = 2 ? d(s, t)/2. In particular, (s|t) w ? 1. As a result

, 2 ? {3/2, 2}. In particular, if d(s, t) = 2 then (s|t) w = 1, and so, we are done because s t , t s ? N [w] and w is simplicial. Otherwise, d(s, t) = 1, and so, (s|t) w = 2. In particular, s t = s and s t

, We have (s|t) w = 3 ? d(s, t)/2 ? {2, 5/2}. In particular

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