YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalSequentialSinglestart(A,B) -> evalSequentialSingleentryin(A,B) True (1,1) 1. evalSequentialSingleentryin(A,B) -> evalSequentialSinglebb1in(0,B) True (?,1) 2. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb5in(A,B) [A >= 0 && A >= B] (?,1) 3. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb2in(A,B) [A >= 0 && B >= 1 + A] (?,1) 4. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + C] (?,1) 5. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1] (?,1) 6. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebb5in(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (?,1) 7. evalSequentialSinglebbin(A,B) -> evalSequentialSinglebb1in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (?,1) 8. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglebb4in(A,B) [A >= 0 && B >= 1 + A] (?,1) 9. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglereturnin(A,B) [A >= 0 && A >= B] (?,1) 10. evalSequentialSinglebb4in(A,B) -> evalSequentialSinglebb5in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (?,1) 11. evalSequentialSinglereturnin(A,B) -> evalSequentialSinglestop(A,B) [A + -1*B >= 0 && A >= 0] (?,1) Signature: {(evalSequentialSinglebb1in,2) ;(evalSequentialSinglebb2in,2) ;(evalSequentialSinglebb4in,2) ;(evalSequentialSinglebb5in,2) ;(evalSequentialSinglebbin,2) ;(evalSequentialSingleentryin,2) ;(evalSequentialSinglereturnin,2) ;(evalSequentialSinglestart,2) ;(evalSequentialSinglestop,2)} Flow Graph: [0->{1},1->{2,3},2->{8,9},3->{4,5,6},4->{7},5->{7},6->{8,9},7->{2,3},8->{10},9->{11},10->{8,9},11->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalSequentialSinglestart(A,B) -> evalSequentialSingleentryin(A,B) True (1,1) 1. evalSequentialSingleentryin(A,B) -> evalSequentialSinglebb1in(0,B) True (1,1) 2. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb5in(A,B) [A >= 0 && A >= B] (1,1) 3. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb2in(A,B) [A >= 0 && B >= 1 + A] (?,1) 4. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + C] (?,1) 5. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1] (?,1) 6. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebb5in(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (1,1) 7. evalSequentialSinglebbin(A,B) -> evalSequentialSinglebb1in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (?,1) 8. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglebb4in(A,B) [A >= 0 && B >= 1 + A] (?,1) 9. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglereturnin(A,B) [A >= 0 && A >= B] (1,1) 10. evalSequentialSinglebb4in(A,B) -> evalSequentialSinglebb5in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (?,1) 11. evalSequentialSinglereturnin(A,B) -> evalSequentialSinglestop(A,B) [A + -1*B >= 0 && A >= 0] (1,1) Signature: {(evalSequentialSinglebb1in,2) ;(evalSequentialSinglebb2in,2) ;(evalSequentialSinglebb4in,2) ;(evalSequentialSinglebb5in,2) ;(evalSequentialSinglebbin,2) ;(evalSequentialSingleentryin,2) ;(evalSequentialSinglereturnin,2) ;(evalSequentialSinglestart,2) ;(evalSequentialSinglestop,2)} Flow Graph: [0->{1},1->{2,3},2->{8,9},3->{4,5,6},4->{7},5->{7},6->{8,9},7->{2,3},8->{10},9->{11},10->{8,9},11->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(2,8),(6,9)] * Step 3: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalSequentialSinglestart(A,B) -> evalSequentialSingleentryin(A,B) True (1,1) 1. evalSequentialSingleentryin(A,B) -> evalSequentialSinglebb1in(0,B) True (1,1) 2. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb5in(A,B) [A >= 0 && A >= B] (1,1) 3. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb2in(A,B) [A >= 0 && B >= 1 + A] (?,1) 4. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + C] (?,1) 5. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1] (?,1) 6. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebb5in(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (1,1) 7. evalSequentialSinglebbin(A,B) -> evalSequentialSinglebb1in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (?,1) 8. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglebb4in(A,B) [A >= 0 && B >= 1 + A] (?,1) 9. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglereturnin(A,B) [A >= 0 && A >= B] (1,1) 10. evalSequentialSinglebb4in(A,B) -> evalSequentialSinglebb5in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (?,1) 11. evalSequentialSinglereturnin(A,B) -> evalSequentialSinglestop(A,B) [A + -1*B >= 0 && A >= 0] (1,1) Signature: {(evalSequentialSinglebb1in,2) ;(evalSequentialSinglebb2in,2) ;(evalSequentialSinglebb4in,2) ;(evalSequentialSinglebb5in,2) ;(evalSequentialSinglebbin,2) ;(evalSequentialSingleentryin,2) ;(evalSequentialSinglereturnin,2) ;(evalSequentialSinglestart,2) ;(evalSequentialSinglestop,2)} Flow Graph: [0->{1},1->{2,3},2->{9},3->{4,5,6},4->{7},5->{7},6->{8},7->{2,3},8->{10},9->{11},10->{8,9},11->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(evalSequentialSinglebb1in) = x2 p(evalSequentialSinglebb2in) = x2 p(evalSequentialSinglebb4in) = -1*x1 + x2 p(evalSequentialSinglebb5in) = -1*x1 + x2 p(evalSequentialSinglebbin) = x2 p(evalSequentialSingleentryin) = x2 p(evalSequentialSinglereturnin) = -1*x1 + x2 p(evalSequentialSinglestart) = x2 p(evalSequentialSinglestop) = -1*x1 + x2 Following rules are strictly oriented: [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] ==> evalSequentialSinglebb4in(A,B) = -1*A + B > -1 + -1*A + B = evalSequentialSinglebb5in(1 + A,B) Following rules are weakly oriented: True ==> evalSequentialSinglestart(A,B) = B >= B = evalSequentialSingleentryin(A,B) True ==> evalSequentialSingleentryin(A,B) = B >= B = evalSequentialSinglebb1in(0,B) [A >= 0 && A >= B] ==> evalSequentialSinglebb1in(A,B) = B >= -1*A + B = evalSequentialSinglebb5in(A,B) [A >= 0 && B >= 1 + A] ==> evalSequentialSinglebb1in(A,B) = B >= B = evalSequentialSinglebb2in(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + C] ==> evalSequentialSinglebb2in(A,B) = B >= B = evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1] ==> evalSequentialSinglebb2in(A,B) = B >= B = evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] ==> evalSequentialSinglebb2in(A,B) = B >= -1*A + B = evalSequentialSinglebb5in(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] ==> evalSequentialSinglebbin(A,B) = B >= B = evalSequentialSinglebb1in(1 + A,B) [A >= 0 && B >= 1 + A] ==> evalSequentialSinglebb5in(A,B) = -1*A + B >= -1*A + B = evalSequentialSinglebb4in(A,B) [A >= 0 && A >= B] ==> evalSequentialSinglebb5in(A,B) = -1*A + B >= -1*A + B = evalSequentialSinglereturnin(A,B) [A + -1*B >= 0 && A >= 0] ==> evalSequentialSinglereturnin(A,B) = -1*A + B >= -1*A + B = evalSequentialSinglestop(A,B) * Step 4: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalSequentialSinglestart(A,B) -> evalSequentialSingleentryin(A,B) True (1,1) 1. evalSequentialSingleentryin(A,B) -> evalSequentialSinglebb1in(0,B) True (1,1) 2. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb5in(A,B) [A >= 0 && A >= B] (1,1) 3. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb2in(A,B) [A >= 0 && B >= 1 + A] (?,1) 4. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + C] (?,1) 5. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1] (?,1) 6. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebb5in(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (1,1) 7. evalSequentialSinglebbin(A,B) -> evalSequentialSinglebb1in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (?,1) 8. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglebb4in(A,B) [A >= 0 && B >= 1 + A] (?,1) 9. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglereturnin(A,B) [A >= 0 && A >= B] (1,1) 10. evalSequentialSinglebb4in(A,B) -> evalSequentialSinglebb5in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (B,1) 11. evalSequentialSinglereturnin(A,B) -> evalSequentialSinglestop(A,B) [A + -1*B >= 0 && A >= 0] (1,1) Signature: {(evalSequentialSinglebb1in,2) ;(evalSequentialSinglebb2in,2) ;(evalSequentialSinglebb4in,2) ;(evalSequentialSinglebb5in,2) ;(evalSequentialSinglebbin,2) ;(evalSequentialSingleentryin,2) ;(evalSequentialSinglereturnin,2) ;(evalSequentialSinglestart,2) ;(evalSequentialSinglestop,2)} Flow Graph: [0->{1},1->{2,3},2->{9},3->{4,5,6},4->{7},5->{7},6->{8},7->{2,3},8->{10},9->{11},10->{8,9},11->{}] + Applied Processor: KnowledgePropagation + Details: We propagate bounds from predecessors. * Step 5: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalSequentialSinglestart(A,B) -> evalSequentialSingleentryin(A,B) True (1,1) 1. evalSequentialSingleentryin(A,B) -> evalSequentialSinglebb1in(0,B) True (1,1) 2. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb5in(A,B) [A >= 0 && A >= B] (1,1) 3. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb2in(A,B) [A >= 0 && B >= 1 + A] (?,1) 4. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + C] (?,1) 5. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1] (?,1) 6. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebb5in(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (1,1) 7. evalSequentialSinglebbin(A,B) -> evalSequentialSinglebb1in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (?,1) 8. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglebb4in(A,B) [A >= 0 && B >= 1 + A] (1 + B,1) 9. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglereturnin(A,B) [A >= 0 && A >= B] (1,1) 10. evalSequentialSinglebb4in(A,B) -> evalSequentialSinglebb5in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (B,1) 11. evalSequentialSinglereturnin(A,B) -> evalSequentialSinglestop(A,B) [A + -1*B >= 0 && A >= 0] (1,1) Signature: {(evalSequentialSinglebb1in,2) ;(evalSequentialSinglebb2in,2) ;(evalSequentialSinglebb4in,2) ;(evalSequentialSinglebb5in,2) ;(evalSequentialSinglebbin,2) ;(evalSequentialSingleentryin,2) ;(evalSequentialSinglereturnin,2) ;(evalSequentialSinglestart,2) ;(evalSequentialSinglestop,2)} Flow Graph: [0->{1},1->{2,3},2->{9},3->{4,5,6},4->{7},5->{7},6->{8},7->{2,3},8->{10},9->{11},10->{8,9},11->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(evalSequentialSinglebb1in) = -1*x1 + x2 p(evalSequentialSinglebb2in) = -1*x1 + x2 p(evalSequentialSinglebb4in) = -1*x1 + x2 p(evalSequentialSinglebb5in) = -1*x1 + x2 p(evalSequentialSinglebbin) = -1*x1 + x2 p(evalSequentialSingleentryin) = x2 p(evalSequentialSinglereturnin) = -1*x1 + x2 p(evalSequentialSinglestart) = x2 p(evalSequentialSinglestop) = -1*x1 + x2 Following rules are strictly oriented: [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] ==> evalSequentialSinglebbin(A,B) = -1*A + B > -1 + -1*A + B = evalSequentialSinglebb1in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] ==> evalSequentialSinglebb4in(A,B) = -1*A + B > -1 + -1*A + B = evalSequentialSinglebb5in(1 + A,B) Following rules are weakly oriented: True ==> evalSequentialSinglestart(A,B) = B >= B = evalSequentialSingleentryin(A,B) True ==> evalSequentialSingleentryin(A,B) = B >= B = evalSequentialSinglebb1in(0,B) [A >= 0 && A >= B] ==> evalSequentialSinglebb1in(A,B) = -1*A + B >= -1*A + B = evalSequentialSinglebb5in(A,B) [A >= 0 && B >= 1 + A] ==> evalSequentialSinglebb1in(A,B) = -1*A + B >= -1*A + B = evalSequentialSinglebb2in(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + C] ==> evalSequentialSinglebb2in(A,B) = -1*A + B >= -1*A + B = evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1] ==> evalSequentialSinglebb2in(A,B) = -1*A + B >= -1*A + B = evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] ==> evalSequentialSinglebb2in(A,B) = -1*A + B >= -1*A + B = evalSequentialSinglebb5in(A,B) [A >= 0 && B >= 1 + A] ==> evalSequentialSinglebb5in(A,B) = -1*A + B >= -1*A + B = evalSequentialSinglebb4in(A,B) [A >= 0 && A >= B] ==> evalSequentialSinglebb5in(A,B) = -1*A + B >= -1*A + B = evalSequentialSinglereturnin(A,B) [A + -1*B >= 0 && A >= 0] ==> evalSequentialSinglereturnin(A,B) = -1*A + B >= -1*A + B = evalSequentialSinglestop(A,B) * Step 6: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalSequentialSinglestart(A,B) -> evalSequentialSingleentryin(A,B) True (1,1) 1. evalSequentialSingleentryin(A,B) -> evalSequentialSinglebb1in(0,B) True (1,1) 2. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb5in(A,B) [A >= 0 && A >= B] (1,1) 3. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb2in(A,B) [A >= 0 && B >= 1 + A] (?,1) 4. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + C] (?,1) 5. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1] (?,1) 6. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebb5in(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (1,1) 7. evalSequentialSinglebbin(A,B) -> evalSequentialSinglebb1in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (B,1) 8. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglebb4in(A,B) [A >= 0 && B >= 1 + A] (1 + B,1) 9. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglereturnin(A,B) [A >= 0 && A >= B] (1,1) 10. evalSequentialSinglebb4in(A,B) -> evalSequentialSinglebb5in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (B,1) 11. evalSequentialSinglereturnin(A,B) -> evalSequentialSinglestop(A,B) [A + -1*B >= 0 && A >= 0] (1,1) Signature: {(evalSequentialSinglebb1in,2) ;(evalSequentialSinglebb2in,2) ;(evalSequentialSinglebb4in,2) ;(evalSequentialSinglebb5in,2) ;(evalSequentialSinglebbin,2) ;(evalSequentialSingleentryin,2) ;(evalSequentialSinglereturnin,2) ;(evalSequentialSinglestart,2) ;(evalSequentialSinglestop,2)} Flow Graph: [0->{1},1->{2,3},2->{9},3->{4,5,6},4->{7},5->{7},6->{8},7->{2,3},8->{10},9->{11},10->{8,9},11->{}] + Applied Processor: KnowledgePropagation + Details: We propagate bounds from predecessors. * Step 7: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalSequentialSinglestart(A,B) -> evalSequentialSingleentryin(A,B) True (1,1) 1. evalSequentialSingleentryin(A,B) -> evalSequentialSinglebb1in(0,B) True (1,1) 2. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb5in(A,B) [A >= 0 && A >= B] (1,1) 3. evalSequentialSinglebb1in(A,B) -> evalSequentialSinglebb2in(A,B) [A >= 0 && B >= 1 + A] (1 + B,1) 4. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + C] (1 + B,1) 5. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebbin(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= 1] (1 + B,1) 6. evalSequentialSinglebb2in(A,B) -> evalSequentialSinglebb5in(A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (1,1) 7. evalSequentialSinglebbin(A,B) -> evalSequentialSinglebb1in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (B,1) 8. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglebb4in(A,B) [A >= 0 && B >= 1 + A] (1 + B,1) 9. evalSequentialSinglebb5in(A,B) -> evalSequentialSinglereturnin(A,B) [A >= 0 && A >= B] (1,1) 10. evalSequentialSinglebb4in(A,B) -> evalSequentialSinglebb5in(1 + A,B) [-1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] (B,1) 11. evalSequentialSinglereturnin(A,B) -> evalSequentialSinglestop(A,B) [A + -1*B >= 0 && A >= 0] (1,1) Signature: {(evalSequentialSinglebb1in,2) ;(evalSequentialSinglebb2in,2) ;(evalSequentialSinglebb4in,2) ;(evalSequentialSinglebb5in,2) ;(evalSequentialSinglebbin,2) ;(evalSequentialSingleentryin,2) ;(evalSequentialSinglereturnin,2) ;(evalSequentialSinglestart,2) ;(evalSequentialSinglestop,2)} Flow Graph: [0->{1},1->{2,3},2->{9},3->{4,5,6},4->{7},5->{7},6->{8},7->{2,3},8->{10},9->{11},10->{8,9},11->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: The problem is already solved. YES(?,O(n^1))