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