YES(?,O(n^1)) * Step 1: UnsatPaths WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. eval0(A,B,C) -> eval1(A,B,C) [A >= 1] (1,1) 1. eval1(A,B,C) -> eval1(A,A + B,C) [-1 + A >= 0 && A >= 1 + B && C >= 1 + A && A >= 1] (?,1) 2. eval1(A,B,C) -> eval1(A,B,-1*A + B) [-1 + A >= 0 && A >= 1 + B && C >= 1 + A && A >= 1] (?,1) Signature: {(eval0,3);(eval1,3)} Flow Graph: [0->{1,2},1->{1,2},2->{1,2}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(2,1),(2,2)] * Step 2: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. eval0(A,B,C) -> eval1(A,B,C) [A >= 1] (1,1) 1. eval1(A,B,C) -> eval1(A,A + B,C) [-1 + A >= 0 && A >= 1 + B && C >= 1 + A && A >= 1] (?,1) 2. eval1(A,B,C) -> eval1(A,B,-1*A + B) [-1 + A >= 0 && A >= 1 + B && C >= 1 + A && A >= 1] (?,1) Signature: {(eval0,3);(eval1,3)} Flow Graph: [0->{1,2},1->{1,2},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. eval0(A,B,C) -> eval1(A,B,C) [A >= 1] (1,1) 1. eval1(A,B,C) -> eval1(A,A + B,C) [-1 + A >= 0 && A >= 1 + B && C >= 1 + A && A >= 1] (?,1) 2. eval1(A,B,C) -> eval1(A,B,-1*A + B) [-1 + A >= 0 && A >= 1 + B && C >= 1 + A && A >= 1] (1,1) Signature: {(eval0,3);(eval1,3)} Flow Graph: [0->{1,2},1->{1,2},2->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(eval0) = x1 + -1*x2 p(eval1) = x1 + -1*x2 Following rules are strictly oriented: [-1 + A >= 0 && A >= 1 + B && C >= 1 + A && A >= 1] ==> eval1(A,B,C) = A + -1*B > -1*B = eval1(A,A + B,C) Following rules are weakly oriented: [A >= 1] ==> eval0(A,B,C) = A + -1*B >= A + -1*B = eval1(A,B,C) [-1 + A >= 0 && A >= 1 + B && C >= 1 + A && A >= 1] ==> eval1(A,B,C) = A + -1*B >= A + -1*B = eval1(A,B,-1*A + B) * Step 4: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. eval0(A,B,C) -> eval1(A,B,C) [A >= 1] (1,1) 1. eval1(A,B,C) -> eval1(A,A + B,C) [-1 + A >= 0 && A >= 1 + B && C >= 1 + A && A >= 1] (A + B,1) 2. eval1(A,B,C) -> eval1(A,B,-1*A + B) [-1 + A >= 0 && A >= 1 + B && C >= 1 + A && A >= 1] (1,1) Signature: {(eval0,3);(eval1,3)} Flow Graph: [0->{1,2},1->{1,2},2->{}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))