YES(?,O(n^1)) * Step 1: UnsatPaths WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f1(A,B,C) -> f300(A,B,D) [A >= B && A >= 1 + B] (?,1) 1. f1(A,B,C) -> f1(1 + A,A,C) [B >= D && A = B] (?,1) 2. f1(A,B,C) -> f1(1 + A,B,C) [B >= 1 + A] (?,1) 3. f2(A,B,C) -> f1(A,B,C) True (1,1) Signature: {(f1,3);(f2,3);(f300,3)} Flow Graph: [0->{},1->{0,1,2},2->{0,1,2},3->{0,1,2}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(1,1),(1,2),(2,0)] * Step 2: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f1(A,B,C) -> f300(A,B,D) [A >= B && A >= 1 + B] (?,1) 1. f1(A,B,C) -> f1(1 + A,A,C) [B >= D && A = B] (?,1) 2. f1(A,B,C) -> f1(1 + A,B,C) [B >= 1 + A] (?,1) 3. f2(A,B,C) -> f1(A,B,C) True (1,1) Signature: {(f1,3);(f2,3);(f300,3)} Flow Graph: [0->{},1->{0},2->{1,2},3->{0,1,2}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 3: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f1(A,B,C) -> f300(A,B,D) [A >= B && A >= 1 + B] (1,1) 1. f1(A,B,C) -> f1(1 + A,A,C) [B >= D && A = B] (1,1) 2. f1(A,B,C) -> f1(1 + A,B,C) [B >= 1 + A] (?,1) 3. f2(A,B,C) -> f1(A,B,C) True (1,1) Signature: {(f1,3);(f2,3);(f300,3)} Flow Graph: [0->{},1->{0},2->{1,2},3->{0,1,2}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = -1*x1 + x2 p(f2) = -1*x1 + x2 p(f300) = -1*x1 + x2 Following rules are strictly oriented: [B >= 1 + A] ==> f1(A,B,C) = -1*A + B > -1 + -1*A + B = f1(1 + A,B,C) Following rules are weakly oriented: [A >= B && A >= 1 + B] ==> f1(A,B,C) = -1*A + B >= -1*A + B = f300(A,B,D) [B >= D && A = B] ==> f1(A,B,C) = -1*A + B >= -1 = f1(1 + A,A,C) True ==> f2(A,B,C) = -1*A + B >= -1*A + B = f1(A,B,C) * Step 4: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f1(A,B,C) -> f300(A,B,D) [A >= B && A >= 1 + B] (1,1) 1. f1(A,B,C) -> f1(1 + A,A,C) [B >= D && A = B] (1,1) 2. f1(A,B,C) -> f1(1 + A,B,C) [B >= 1 + A] (A + B,1) 3. f2(A,B,C) -> f1(A,B,C) True (1,1) Signature: {(f1,3);(f2,3);(f300,3)} Flow Graph: [0->{},1->{0},2->{1,2},3->{0,1,2}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))