YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f2(A,B) -> f300(-1 + A,C) [0 >= A] (?,1) 1. f2(A,B) -> f2(-1 + A,B) [A >= 1] (?,1) 2. f3(A,B) -> f2(A,B) True (1,1) Signature: {(f2,2);(f3,2);(f300,2)} Flow Graph: [0->{},1->{0,1},2->{0,1}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f2(A,B) -> f300(-1 + A,C) [0 >= A] (1,1) 1. f2(A,B) -> f2(-1 + A,B) [A >= 1] (?,1) 2. f3(A,B) -> f2(A,B) True (1,1) Signature: {(f2,2);(f3,2);(f300,2)} Flow Graph: [0->{},1->{0,1},2->{0,1}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f2) = x1 p(f3) = x1 p(f300) = 1 + x1 Following rules are strictly oriented: [A >= 1] ==> f2(A,B) = A > -1 + A = f2(-1 + A,B) Following rules are weakly oriented: [0 >= A] ==> f2(A,B) = A >= A = f300(-1 + A,C) True ==> f3(A,B) = A >= A = f2(A,B) * Step 3: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f2(A,B) -> f300(-1 + A,C) [0 >= A] (1,1) 1. f2(A,B) -> f2(-1 + A,B) [A >= 1] (A,1) 2. f3(A,B) -> f2(A,B) True (1,1) Signature: {(f2,2);(f3,2);(f300,2)} Flow Graph: [0->{},1->{0,1},2->{0,1}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))