YES(?,O(1)) * Step 1: UnsatPaths WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f300(A,B,C,D,E) -> f300(-1 + A,B,C,D,E) [1000 + -1*A >= 0 && A >= 101 && 9 >= B] (?,1) 1. f300(A,B,C,D,E) -> f2(A,B,0,0,0) [1000 + -1*A >= 0 && 100 >= A && 9 >= B] (?,1) 2. f300(A,B,C,D,E) -> f2(A,B,0,0,0) [1000 + -1*A >= 0 && B >= 10] (?,1) 3. f1(A,B,C,D,E) -> f300(1000,B,C,D,E) True (1,1) Signature: {(f1,5);(f2,5);(f300,5)} Flow Graph: [0->{0,1,2},1->{},2->{},3->{0,1,2}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(0,2),(3,1)] * Step 2: TrivialSCCs WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f300(A,B,C,D,E) -> f300(-1 + A,B,C,D,E) [1000 + -1*A >= 0 && A >= 101 && 9 >= B] (?,1) 1. f300(A,B,C,D,E) -> f2(A,B,0,0,0) [1000 + -1*A >= 0 && 100 >= A && 9 >= B] (?,1) 2. f300(A,B,C,D,E) -> f2(A,B,0,0,0) [1000 + -1*A >= 0 && B >= 10] (?,1) 3. f1(A,B,C,D,E) -> f300(1000,B,C,D,E) True (1,1) Signature: {(f1,5);(f2,5);(f300,5)} Flow Graph: [0->{0,1},1->{},2->{},3->{0,2}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 3: PolyRank WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f300(A,B,C,D,E) -> f300(-1 + A,B,C,D,E) [1000 + -1*A >= 0 && A >= 101 && 9 >= B] (?,1) 1. f300(A,B,C,D,E) -> f2(A,B,0,0,0) [1000 + -1*A >= 0 && 100 >= A && 9 >= B] (1,1) 2. f300(A,B,C,D,E) -> f2(A,B,0,0,0) [1000 + -1*A >= 0 && B >= 10] (1,1) 3. f1(A,B,C,D,E) -> f300(1000,B,C,D,E) True (1,1) Signature: {(f1,5);(f2,5);(f300,5)} Flow Graph: [0->{0,1},1->{},2->{},3->{0,2}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = 1000 p(f2) = x1 p(f300) = x1 Following rules are strictly oriented: [1000 + -1*A >= 0 && A >= 101 && 9 >= B] ==> f300(A,B,C,D,E) = A > -1 + A = f300(-1 + A,B,C,D,E) Following rules are weakly oriented: [1000 + -1*A >= 0 && 100 >= A && 9 >= B] ==> f300(A,B,C,D,E) = A >= A = f2(A,B,0,0,0) [1000 + -1*A >= 0 && B >= 10] ==> f300(A,B,C,D,E) = A >= A = f2(A,B,0,0,0) True ==> f1(A,B,C,D,E) = 1000 >= 1000 = f300(1000,B,C,D,E) * Step 4: KnowledgePropagation WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f300(A,B,C,D,E) -> f300(-1 + A,B,C,D,E) [1000 + -1*A >= 0 && A >= 101 && 9 >= B] (1000,1) 1. f300(A,B,C,D,E) -> f2(A,B,0,0,0) [1000 + -1*A >= 0 && 100 >= A && 9 >= B] (1,1) 2. f300(A,B,C,D,E) -> f2(A,B,0,0,0) [1000 + -1*A >= 0 && B >= 10] (1,1) 3. f1(A,B,C,D,E) -> f300(1000,B,C,D,E) True (1,1) Signature: {(f1,5);(f2,5);(f300,5)} Flow Graph: [0->{0,1},1->{},2->{},3->{0,2}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(1))