YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalfstart(A,B) -> evalfentryin(A,B) True (1,1) 1. evalfentryin(A,B) -> evalfbb1in(B,A) True (?,1) 2. evalfbb1in(A,B) -> evalfbbin(A,B) [A >= B] (?,1) 3. evalfbb1in(A,B) -> evalfreturnin(A,B) [B >= 1 + A] (?,1) 4. evalfbbin(A,B) -> evalfbb1in(A,1 + B) [A + -1*B >= 0] (?,1) 5. evalfreturnin(A,B) -> evalfstop(A,B) [-1 + -1*A + B >= 0] (?,1) Signature: {(evalfbb1in,2);(evalfbbin,2);(evalfentryin,2);(evalfreturnin,2);(evalfstart,2);(evalfstop,2)} Flow Graph: [0->{1},1->{2,3},2->{4},3->{5},4->{2,3},5->{}] + 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. evalfstart(A,B) -> evalfentryin(A,B) True (1,1) 1. evalfentryin(A,B) -> evalfbb1in(B,A) True (1,1) 2. evalfbb1in(A,B) -> evalfbbin(A,B) [A >= B] (?,1) 3. evalfbb1in(A,B) -> evalfreturnin(A,B) [B >= 1 + A] (1,1) 4. evalfbbin(A,B) -> evalfbb1in(A,1 + B) [A + -1*B >= 0] (?,1) 5. evalfreturnin(A,B) -> evalfstop(A,B) [-1 + -1*A + B >= 0] (1,1) Signature: {(evalfbb1in,2);(evalfbbin,2);(evalfentryin,2);(evalfreturnin,2);(evalfstart,2);(evalfstop,2)} Flow Graph: [0->{1},1->{2,3},2->{4},3->{5},4->{2,3},5->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(evalfbb1in) = 1 + x1 + -1*x2 p(evalfbbin) = 1 + x1 + -1*x2 p(evalfentryin) = 1 + -1*x1 + x2 p(evalfreturnin) = 1 + x1 + -1*x2 p(evalfstart) = 1 + -1*x1 + x2 p(evalfstop) = 1 + x1 + -1*x2 Following rules are strictly oriented: [A + -1*B >= 0] ==> evalfbbin(A,B) = 1 + A + -1*B > A + -1*B = evalfbb1in(A,1 + B) Following rules are weakly oriented: True ==> evalfstart(A,B) = 1 + -1*A + B >= 1 + -1*A + B = evalfentryin(A,B) True ==> evalfentryin(A,B) = 1 + -1*A + B >= 1 + -1*A + B = evalfbb1in(B,A) [A >= B] ==> evalfbb1in(A,B) = 1 + A + -1*B >= 1 + A + -1*B = evalfbbin(A,B) [B >= 1 + A] ==> evalfbb1in(A,B) = 1 + A + -1*B >= 1 + A + -1*B = evalfreturnin(A,B) [-1 + -1*A + B >= 0] ==> evalfreturnin(A,B) = 1 + A + -1*B >= 1 + A + -1*B = evalfstop(A,B) * Step 3: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalfstart(A,B) -> evalfentryin(A,B) True (1,1) 1. evalfentryin(A,B) -> evalfbb1in(B,A) True (1,1) 2. evalfbb1in(A,B) -> evalfbbin(A,B) [A >= B] (?,1) 3. evalfbb1in(A,B) -> evalfreturnin(A,B) [B >= 1 + A] (1,1) 4. evalfbbin(A,B) -> evalfbb1in(A,1 + B) [A + -1*B >= 0] (1 + A + B,1) 5. evalfreturnin(A,B) -> evalfstop(A,B) [-1 + -1*A + B >= 0] (1,1) Signature: {(evalfbb1in,2);(evalfbbin,2);(evalfentryin,2);(evalfreturnin,2);(evalfstart,2);(evalfstop,2)} Flow Graph: [0->{1},1->{2,3},2->{4},3->{5},4->{2,3},5->{}] + Applied Processor: KnowledgePropagation + Details: We propagate bounds from predecessors. * Step 4: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalfstart(A,B) -> evalfentryin(A,B) True (1,1) 1. evalfentryin(A,B) -> evalfbb1in(B,A) True (1,1) 2. evalfbb1in(A,B) -> evalfbbin(A,B) [A >= B] (2 + A + B,1) 3. evalfbb1in(A,B) -> evalfreturnin(A,B) [B >= 1 + A] (1,1) 4. evalfbbin(A,B) -> evalfbb1in(A,1 + B) [A + -1*B >= 0] (1 + A + B,1) 5. evalfreturnin(A,B) -> evalfstop(A,B) [-1 + -1*A + B >= 0] (1,1) Signature: {(evalfbb1in,2);(evalfbbin,2);(evalfentryin,2);(evalfreturnin,2);(evalfstart,2);(evalfstop,2)} Flow Graph: [0->{1},1->{2,3},2->{4},3->{5},4->{2,3},5->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: The problem is already solved. YES(?,O(n^1))