YES(?,O(n^1)) * Step 1: UnsatPaths WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalNestedSinglestart(A,B,C) -> evalNestedSingleentryin(A,B,C) True (1,1) 1. evalNestedSingleentryin(A,B,C) -> evalNestedSinglebb5in(0,B,C) True (?,1) 2. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglebb2in(A,B,A) [B >= 1 + A] (?,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= B] (?,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= B] (?,1) 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [B >= 1 + C] (?,1) 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [0 >= 1 + D] (?,1) 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [D >= 1] (?,1) 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) True (?,1) 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) True (?,1) 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) True (?,1) 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) True (?,1) Signature: {(evalNestedSinglebb1in,3) ;(evalNestedSinglebb2in,3) ;(evalNestedSinglebb3in,3) ;(evalNestedSinglebb4in,3) ;(evalNestedSinglebb5in,3) ;(evalNestedSingleentryin,3) ;(evalNestedSinglereturnin,3) ;(evalNestedSinglestart,3) ;(evalNestedSinglestop,3)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(2,4)] * Step 2: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalNestedSinglestart(A,B,C) -> evalNestedSingleentryin(A,B,C) True (1,1) 1. evalNestedSingleentryin(A,B,C) -> evalNestedSinglebb5in(0,B,C) True (?,1) 2. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglebb2in(A,B,A) [B >= 1 + A] (?,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= B] (?,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= B] (?,1) 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [B >= 1 + C] (?,1) 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [0 >= 1 + D] (?,1) 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [D >= 1] (?,1) 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) True (?,1) 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) True (?,1) 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) True (?,1) 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) True (?,1) Signature: {(evalNestedSinglebb1in,3) ;(evalNestedSinglebb2in,3) ;(evalNestedSinglebb3in,3) ;(evalNestedSinglebb4in,3) ;(evalNestedSinglebb5in,3) ;(evalNestedSingleentryin,3) ;(evalNestedSinglereturnin,3) ;(evalNestedSinglestart,3) ;(evalNestedSinglestop,3)} Flow Graph: [0->{1},1->{2,3},2->{5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}] + 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. evalNestedSinglestart(A,B,C) -> evalNestedSingleentryin(A,B,C) True (1,1) 1. evalNestedSingleentryin(A,B,C) -> evalNestedSinglebb5in(0,B,C) True (1,1) 2. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglebb2in(A,B,A) [B >= 1 + A] (?,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= B] (1,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= B] (?,1) 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [B >= 1 + C] (?,1) 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [0 >= 1 + D] (?,1) 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [D >= 1] (?,1) 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) True (?,1) 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) True (?,1) 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) True (?,1) 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) True (1,1) Signature: {(evalNestedSinglebb1in,3) ;(evalNestedSinglebb2in,3) ;(evalNestedSinglebb3in,3) ;(evalNestedSinglebb4in,3) ;(evalNestedSinglebb5in,3) ;(evalNestedSingleentryin,3) ;(evalNestedSinglereturnin,3) ;(evalNestedSinglestart,3) ;(evalNestedSinglestop,3)} Flow Graph: [0->{1},1->{2,3},2->{5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(evalNestedSinglebb1in) = 1 + x2 + -1*x3 p(evalNestedSinglebb2in) = 2 + x2 + -1*x3 p(evalNestedSinglebb3in) = 1 + x2 + -1*x3 p(evalNestedSinglebb4in) = 1 + x2 + -1*x3 p(evalNestedSinglebb5in) = 2 + -1*x1 + x2 p(evalNestedSingleentryin) = 2 + x2 p(evalNestedSinglereturnin) = 2 + -1*x1 + x2 p(evalNestedSinglestart) = 2 + x2 p(evalNestedSinglestop) = 2 + -1*x1 + x2 Following rules are strictly oriented: [B >= 1 + C] ==> evalNestedSinglebb2in(A,B,C) = 2 + B + -1*C > 1 + B + -1*C = evalNestedSinglebb3in(A,B,C) Following rules are weakly oriented: True ==> evalNestedSinglestart(A,B,C) = 2 + B >= 2 + B = evalNestedSingleentryin(A,B,C) True ==> evalNestedSingleentryin(A,B,C) = 2 + B >= 2 + B = evalNestedSinglebb5in(0,B,C) [B >= 1 + A] ==> evalNestedSinglebb5in(A,B,C) = 2 + -1*A + B >= 2 + -1*A + B = evalNestedSinglebb2in(A,B,A) [A >= B] ==> evalNestedSinglebb5in(A,B,C) = 2 + -1*A + B >= 2 + -1*A + B = evalNestedSinglereturnin(A,B,C) [C >= B] ==> evalNestedSinglebb2in(A,B,C) = 2 + B + -1*C >= 1 + B + -1*C = evalNestedSinglebb4in(A,B,C) [0 >= 1 + D] ==> evalNestedSinglebb3in(A,B,C) = 1 + B + -1*C >= 1 + B + -1*C = evalNestedSinglebb1in(A,B,C) [D >= 1] ==> evalNestedSinglebb3in(A,B,C) = 1 + B + -1*C >= 1 + B + -1*C = evalNestedSinglebb1in(A,B,C) True ==> evalNestedSinglebb3in(A,B,C) = 1 + B + -1*C >= 1 + B + -1*C = evalNestedSinglebb4in(A,B,C) True ==> evalNestedSinglebb1in(A,B,C) = 1 + B + -1*C >= 1 + B + -1*C = evalNestedSinglebb2in(A,B,1 + C) True ==> evalNestedSinglebb4in(A,B,C) = 1 + B + -1*C >= 1 + B + -1*C = evalNestedSinglebb5in(1 + C,B,C) True ==> evalNestedSinglereturnin(A,B,C) = 2 + -1*A + B >= 2 + -1*A + B = evalNestedSinglestop(A,B,C) * Step 4: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalNestedSinglestart(A,B,C) -> evalNestedSingleentryin(A,B,C) True (1,1) 1. evalNestedSingleentryin(A,B,C) -> evalNestedSinglebb5in(0,B,C) True (1,1) 2. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglebb2in(A,B,A) [B >= 1 + A] (?,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= B] (1,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= B] (?,1) 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [B >= 1 + C] (2 + B,1) 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [0 >= 1 + D] (?,1) 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [D >= 1] (?,1) 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) True (?,1) 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) True (?,1) 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) True (?,1) 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) True (1,1) Signature: {(evalNestedSinglebb1in,3) ;(evalNestedSinglebb2in,3) ;(evalNestedSinglebb3in,3) ;(evalNestedSinglebb4in,3) ;(evalNestedSinglebb5in,3) ;(evalNestedSingleentryin,3) ;(evalNestedSinglereturnin,3) ;(evalNestedSinglestart,3) ;(evalNestedSinglestop,3)} Flow Graph: [0->{1},1->{2,3},2->{5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}] + Applied Processor: KnowledgePropagation + Details: We propagate bounds from predecessors. * Step 5: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalNestedSinglestart(A,B,C) -> evalNestedSingleentryin(A,B,C) True (1,1) 1. evalNestedSingleentryin(A,B,C) -> evalNestedSinglebb5in(0,B,C) True (1,1) 2. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglebb2in(A,B,A) [B >= 1 + A] (?,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= B] (1,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= B] (4 + 2*B,1) 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [B >= 1 + C] (2 + B,1) 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [0 >= 1 + D] (2 + B,1) 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [D >= 1] (2 + B,1) 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) True (2 + B,1) 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) True (4 + 2*B,1) 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) True (6 + 3*B,1) 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) True (1,1) Signature: {(evalNestedSinglebb1in,3) ;(evalNestedSinglebb2in,3) ;(evalNestedSinglebb3in,3) ;(evalNestedSinglebb4in,3) ;(evalNestedSinglebb5in,3) ;(evalNestedSingleentryin,3) ;(evalNestedSinglereturnin,3) ;(evalNestedSinglestart,3) ;(evalNestedSinglestop,3)} Flow Graph: [0->{1},1->{2,3},2->{5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(evalNestedSinglebb1in) = x2 + -1*x3 p(evalNestedSinglebb2in) = 1 + x2 + -1*x3 p(evalNestedSinglebb3in) = 1 + x2 + -1*x3 p(evalNestedSinglebb4in) = 1 + x2 + -1*x3 p(evalNestedSinglebb5in) = 2 + -1*x1 + x2 p(evalNestedSingleentryin) = 2 + x2 p(evalNestedSinglereturnin) = 2 + -1*x1 + x2 p(evalNestedSinglestart) = 2 + x2 p(evalNestedSinglestop) = 2 + -1*x1 + x2 Following rules are strictly oriented: [B >= 1 + A] ==> evalNestedSinglebb5in(A,B,C) = 2 + -1*A + B > 1 + -1*A + B = evalNestedSinglebb2in(A,B,A) Following rules are weakly oriented: True ==> evalNestedSinglestart(A,B,C) = 2 + B >= 2 + B = evalNestedSingleentryin(A,B,C) True ==> evalNestedSingleentryin(A,B,C) = 2 + B >= 2 + B = evalNestedSinglebb5in(0,B,C) [A >= B] ==> evalNestedSinglebb5in(A,B,C) = 2 + -1*A + B >= 2 + -1*A + B = evalNestedSinglereturnin(A,B,C) [C >= B] ==> evalNestedSinglebb2in(A,B,C) = 1 + B + -1*C >= 1 + B + -1*C = evalNestedSinglebb4in(A,B,C) [B >= 1 + C] ==> evalNestedSinglebb2in(A,B,C) = 1 + B + -1*C >= 1 + B + -1*C = evalNestedSinglebb3in(A,B,C) [0 >= 1 + D] ==> evalNestedSinglebb3in(A,B,C) = 1 + B + -1*C >= B + -1*C = evalNestedSinglebb1in(A,B,C) [D >= 1] ==> evalNestedSinglebb3in(A,B,C) = 1 + B + -1*C >= B + -1*C = evalNestedSinglebb1in(A,B,C) True ==> evalNestedSinglebb3in(A,B,C) = 1 + B + -1*C >= 1 + B + -1*C = evalNestedSinglebb4in(A,B,C) True ==> evalNestedSinglebb1in(A,B,C) = B + -1*C >= B + -1*C = evalNestedSinglebb2in(A,B,1 + C) True ==> evalNestedSinglebb4in(A,B,C) = 1 + B + -1*C >= 1 + B + -1*C = evalNestedSinglebb5in(1 + C,B,C) True ==> evalNestedSinglereturnin(A,B,C) = 2 + -1*A + B >= 2 + -1*A + B = evalNestedSinglestop(A,B,C) * Step 6: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalNestedSinglestart(A,B,C) -> evalNestedSingleentryin(A,B,C) True (1,1) 1. evalNestedSingleentryin(A,B,C) -> evalNestedSinglebb5in(0,B,C) True (1,1) 2. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglebb2in(A,B,A) [B >= 1 + A] (2 + B,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= B] (1,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= B] (4 + 2*B,1) 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [B >= 1 + C] (2 + B,1) 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [0 >= 1 + D] (2 + B,1) 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [D >= 1] (2 + B,1) 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) True (2 + B,1) 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) True (4 + 2*B,1) 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) True (6 + 3*B,1) 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) True (1,1) Signature: {(evalNestedSinglebb1in,3) ;(evalNestedSinglebb2in,3) ;(evalNestedSinglebb3in,3) ;(evalNestedSinglebb4in,3) ;(evalNestedSinglebb5in,3) ;(evalNestedSingleentryin,3) ;(evalNestedSinglereturnin,3) ;(evalNestedSinglestart,3) ;(evalNestedSinglestop,3)} Flow Graph: [0->{1},1->{2,3},2->{5},3->{11},4->{10},5->{6,7,8},6->{9},7->{9},8->{10},9->{4,5},10->{2,3},11->{}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))