YES(?,O(n^1)) * Step 1: 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) [A >= 0 && B >= 1 + A] (?,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= 0 && A >= B] (?,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= B] 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && B >= 1 + C] 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + D] 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= 1] 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) [A + -1*B >= 0 && A >= 0] (?,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: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: 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 (1,1) 2. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglebb2in(A,B,A) [A >= 0 && B >= 1 + A] (?,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= 0 && A >= B] (1,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= B] 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && B >= 1 + C] 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + D] 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= 1] 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) [A + -1*B >= 0 && A >= 0] (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->{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 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) [A >= 0 && B >= 1 + A] (?,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= 0 && A >= B] (1,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= B] 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && B >= 1 + C] 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + D] 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= 1] 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) [A + -1*B >= 0 && A >= 0] (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*x1 + x2 p(evalNestedSinglebb2in) = -1*x1 + x2 p(evalNestedSinglebb3in) = -1*x1 + x2 p(evalNestedSinglebb4in) = -1*x1 + x2 p(evalNestedSinglebb5in) = -1*x1 + x2 p(evalNestedSingleentryin) = x2 p(evalNestedSinglereturnin) = -1*x1 + x2 p(evalNestedSinglestart) = x2 p(evalNestedSinglestop) = -1*x1 + x2 Following rules are strictly oriented: [C >= 0 ==> && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] evalNestedSinglebb4in(A,B,C) = -1*A + B > -1 + B + -1*C = evalNestedSinglebb5in(1 + C,B,C) Following rules are weakly oriented: True ==> evalNestedSinglestart(A,B,C) = B >= B = evalNestedSingleentryin(A,B,C) True ==> evalNestedSingleentryin(A,B,C) = B >= B = evalNestedSinglebb5in(0,B,C) [A >= 0 && B >= 1 + A] ==> evalNestedSinglebb5in(A,B,C) = -1*A + B >= -1*A + B = evalNestedSinglebb2in(A,B,A) [A >= 0 && A >= B] ==> evalNestedSinglebb5in(A,B,C) = -1*A + B >= -1*A + B = evalNestedSinglereturnin(A,B,C) [C >= 0 ==> && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= B] evalNestedSinglebb2in(A,B,C) = -1*A + B >= -1*A + B = evalNestedSinglebb4in(A,B,C) [C >= 0 ==> && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && B >= 1 + C] evalNestedSinglebb2in(A,B,C) = -1*A + B >= -1*A + B = evalNestedSinglebb3in(A,B,C) [-1 + B + -1*C >= 0 ==> && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + D] evalNestedSinglebb3in(A,B,C) = -1*A + B >= -1*A + B = evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 ==> && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= 1] evalNestedSinglebb3in(A,B,C) = -1*A + B >= -1*A + B = evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 ==> && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] evalNestedSinglebb3in(A,B,C) = -1*A + B >= -1*A + B = evalNestedSinglebb4in(A,B,C) [-1 + B + -1*C >= 0 ==> && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] evalNestedSinglebb1in(A,B,C) = -1*A + B >= -1*A + B = evalNestedSinglebb2in(A,B,1 + C) [A + -1*B >= 0 && A >= 0] ==> evalNestedSinglereturnin(A,B,C) = -1*A + B >= -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) [A >= 0 && B >= 1 + A] (?,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= 0 && A >= B] (1,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= B] 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && B >= 1 + C] 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + D] 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= 1] 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) [C >= 0 (B,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) [A + -1*B >= 0 && A >= 0] (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) [A >= 0 && B >= 1 + A] (1 + B,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= 0 && A >= B] (1,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= B] 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && B >= 1 + C] 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + D] 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= 1] 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) [C >= 0 (B,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) [A + -1*B >= 0 && A >= 0] (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) = x2 + -1*x3 p(evalNestedSinglebb3in) = x2 + -1*x3 p(evalNestedSinglebb4in) = x2 + -1*x3 p(evalNestedSinglebb5in) = -1*x1 + x2 p(evalNestedSingleentryin) = x2 p(evalNestedSinglereturnin) = -1*x1 + x2 p(evalNestedSinglestart) = x2 p(evalNestedSinglestop) = -1*x1 + x2 Following rules are strictly oriented: [-1 + B + -1*C >= 0 ==> && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] evalNestedSinglebb1in(A,B,C) = B + -1*C > -1 + B + -1*C = evalNestedSinglebb2in(A,B,1 + C) Following rules are weakly oriented: True ==> evalNestedSinglestart(A,B,C) = B >= B = evalNestedSingleentryin(A,B,C) True ==> evalNestedSingleentryin(A,B,C) = B >= B = evalNestedSinglebb5in(0,B,C) [A >= 0 && B >= 1 + A] ==> evalNestedSinglebb5in(A,B,C) = -1*A + B >= -1*A + B = evalNestedSinglebb2in(A,B,A) [A >= 0 && A >= B] ==> evalNestedSinglebb5in(A,B,C) = -1*A + B >= -1*A + B = evalNestedSinglereturnin(A,B,C) [C >= 0 ==> && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= B] evalNestedSinglebb2in(A,B,C) = B + -1*C >= B + -1*C = evalNestedSinglebb4in(A,B,C) [C >= 0 ==> && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && B >= 1 + C] evalNestedSinglebb2in(A,B,C) = B + -1*C >= B + -1*C = evalNestedSinglebb3in(A,B,C) [-1 + B + -1*C >= 0 ==> && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + D] evalNestedSinglebb3in(A,B,C) = B + -1*C >= B + -1*C = evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 ==> && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= 1] evalNestedSinglebb3in(A,B,C) = B + -1*C >= B + -1*C = evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 ==> && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] evalNestedSinglebb3in(A,B,C) = B + -1*C >= B + -1*C = evalNestedSinglebb4in(A,B,C) [C >= 0 ==> && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] evalNestedSinglebb4in(A,B,C) = B + -1*C >= -1 + B + -1*C = evalNestedSinglebb5in(1 + C,B,C) [A + -1*B >= 0 && A >= 0] ==> evalNestedSinglereturnin(A,B,C) = -1*A + B >= -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) [A >= 0 && B >= 1 + A] (1 + B,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= 0 && A >= B] (1,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= B] 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && B >= 1 + C] 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + D] 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= 1] 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [-1 + B + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) [-1 + B + -1*C >= 0 (B,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) [C >= 0 (B,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) [A + -1*B >= 0 && A >= 0] (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 7: 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) [A >= 0 && B >= 1 + A] (1 + B,1) 3. evalNestedSinglebb5in(A,B,C) -> evalNestedSinglereturnin(A,B,C) [A >= 0 && A >= B] (1,1) 4. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [C >= 0 (B,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && C >= B] 5. evalNestedSinglebb2in(A,B,C) -> evalNestedSinglebb3in(A,B,C) [C >= 0 (1 + 2*B,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && B >= 1 + C] 6. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (1 + 2*B,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && 0 >= 1 + D] 7. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb1in(A,B,C) [-1 + B + -1*C >= 0 (1 + 2*B,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0 && D >= 1] 8. evalNestedSinglebb3in(A,B,C) -> evalNestedSinglebb4in(A,B,C) [-1 + B + -1*C >= 0 (1 + 2*B,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 9. evalNestedSinglebb1in(A,B,C) -> evalNestedSinglebb2in(A,B,1 + C) [-1 + B + -1*C >= 0 (B,1) && C >= 0 && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 10. evalNestedSinglebb4in(A,B,C) -> evalNestedSinglebb5in(1 + C,B,C) [C >= 0 (B,1) && -1 + B + C >= 0 && A + C >= 0 && -1*A + C >= 0 && -1 + B >= 0 && -1 + A + B >= 0 && -1 + -1*A + B >= 0 && A >= 0] 11. evalNestedSinglereturnin(A,B,C) -> evalNestedSinglestop(A,B,C) [A + -1*B >= 0 && A >= 0] (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: The problem is already solved. YES(?,O(n^1))