YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalfstart(A,B,C) -> evalfentryin(A,B,C) True (1,1) 1. evalfentryin(A,B,C) -> evalfbb3in(B,A,0) [A >= 1 && B >= 1] (?,1) 2. evalfbb3in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && 0 >= B] (?,1) 3. evalfbb3in(A,B,C) -> evalfbb4in(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && B >= 1] (?,1) 4. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + D] 5. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && D >= 1] (?,1) 6. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] (?,1) 7. evalfbbin(A,B,C) -> evalfbb1in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 1 + C] 8. evalfbbin(A,B,C) -> evalfbb3in(A,B,0) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && C >= A] (?,1) 9. evalfbb1in(A,B,C) -> evalfbb3in(A,-1 + B,1 + C) [-1 + A + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] 10. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0] (?,1) Signature: {(evalfbb1in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfbbin,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{2,3},2->{10},3->{4,5,6},4->{7,8},5->{7,8},6->{10},7->{9},8->{2,3},9->{2,3},10->{}] + 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. evalfstart(A,B,C) -> evalfentryin(A,B,C) True (1,1) 1. evalfentryin(A,B,C) -> evalfbb3in(B,A,0) [A >= 1 && B >= 1] (1,1) 2. evalfbb3in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && 0 >= B] (1,1) 3. evalfbb3in(A,B,C) -> evalfbb4in(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && B >= 1] (?,1) 4. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + D] 5. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && D >= 1] (?,1) 6. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] (1,1) 7. evalfbbin(A,B,C) -> evalfbb1in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 1 + C] 8. evalfbbin(A,B,C) -> evalfbb3in(A,B,0) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && C >= A] (?,1) 9. evalfbb1in(A,B,C) -> evalfbb3in(A,-1 + B,1 + C) [-1 + A + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] 10. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0] (1,1) Signature: {(evalfbb1in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfbbin,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{2,3},2->{10},3->{4,5,6},4->{7,8},5->{7,8},6->{10},7->{9},8->{2,3},9->{2,3},10->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(1,2),(8,2)] * Step 3: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalfstart(A,B,C) -> evalfentryin(A,B,C) True (1,1) 1. evalfentryin(A,B,C) -> evalfbb3in(B,A,0) [A >= 1 && B >= 1] (1,1) 2. evalfbb3in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && 0 >= B] (1,1) 3. evalfbb3in(A,B,C) -> evalfbb4in(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && B >= 1] (?,1) 4. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + D] 5. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && D >= 1] (?,1) 6. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] (1,1) 7. evalfbbin(A,B,C) -> evalfbb1in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 1 + C] 8. evalfbbin(A,B,C) -> evalfbb3in(A,B,0) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && C >= A] (?,1) 9. evalfbb1in(A,B,C) -> evalfbb3in(A,-1 + B,1 + C) [-1 + A + -1*C >= 0 (?,1) && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] 10. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0] (1,1) Signature: {(evalfbb1in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfbbin,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{3},2->{10},3->{4,5,6},4->{7,8},5->{7,8},6->{10},7->{9},8->{3},9->{2,3},10->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(evalfbb1in) = x2 p(evalfbb3in) = x2 p(evalfbb4in) = x2 p(evalfbbin) = x2 p(evalfentryin) = x1 p(evalfreturnin) = x2 p(evalfstart) = x1 p(evalfstop) = x2 Following rules are strictly oriented: [-1 + A + -1*C >= 0 ==> && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] evalfbb1in(A,B,C) = B > -1 + B = evalfbb3in(A,-1 + B,1 + C) Following rules are weakly oriented: True ==> evalfstart(A,B,C) = A >= A = evalfentryin(A,B,C) [A >= 1 && B >= 1] ==> evalfentryin(A,B,C) = A >= A = evalfbb3in(B,A,0) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && 0 >= B] ==> evalfbb3in(A,B,C) = B >= B = evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && B >= 1] ==> evalfbb3in(A,B,C) = B >= B = evalfbb4in(A,B,C) [C >= 0 ==> && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + D] evalfbb4in(A,B,C) = B >= B = evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && D >= 1] ==> evalfbb4in(A,B,C) = B >= B = evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] ==> evalfbb4in(A,B,C) = B >= B = evalfreturnin(A,B,C) [C >= 0 ==> && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 1 + C] evalfbbin(A,B,C) = B >= B = evalfbb1in(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && C >= A] ==> evalfbbin(A,B,C) = B >= B = evalfbb3in(A,B,0) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0] ==> evalfreturnin(A,B,C) = B >= B = evalfstop(A,B,C) * Step 4: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalfstart(A,B,C) -> evalfentryin(A,B,C) True (1,1) 1. evalfentryin(A,B,C) -> evalfbb3in(B,A,0) [A >= 1 && B >= 1] (1,1) 2. evalfbb3in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && 0 >= B] (1,1) 3. evalfbb3in(A,B,C) -> evalfbb4in(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && B >= 1] (?,1) 4. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + D] 5. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && D >= 1] (?,1) 6. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] (1,1) 7. evalfbbin(A,B,C) -> evalfbb1in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 1 + C] 8. evalfbbin(A,B,C) -> evalfbb3in(A,B,0) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && C >= A] (?,1) 9. evalfbb1in(A,B,C) -> evalfbb3in(A,-1 + B,1 + C) [-1 + A + -1*C >= 0 (A,1) && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] 10. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0] (1,1) Signature: {(evalfbb1in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfbbin,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{3},2->{10},3->{4,5,6},4->{7,8},5->{7,8},6->{10},7->{9},8->{3},9->{2,3},10->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(evalfbb1in) = x2 + x3 p(evalfbb3in) = x2 + x3 p(evalfbb4in) = x2 + x3 p(evalfbbin) = x2 + x3 p(evalfentryin) = x1 p(evalfreturnin) = x2 + x3 p(evalfstart) = x1 p(evalfstop) = x2 + x3 Following rules are strictly oriented: [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && C >= A] ==> evalfbbin(A,B,C) = B + C > B = evalfbb3in(A,B,0) Following rules are weakly oriented: True ==> evalfstart(A,B,C) = A >= A = evalfentryin(A,B,C) [A >= 1 && B >= 1] ==> evalfentryin(A,B,C) = A >= A = evalfbb3in(B,A,0) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && 0 >= B] ==> evalfbb3in(A,B,C) = B + C >= B + C = evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && B >= 1] ==> evalfbb3in(A,B,C) = B + C >= B + C = evalfbb4in(A,B,C) [C >= 0 ==> && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + D] evalfbb4in(A,B,C) = B + C >= B + C = evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && D >= 1] ==> evalfbb4in(A,B,C) = B + C >= B + C = evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] ==> evalfbb4in(A,B,C) = B + C >= B + C = evalfreturnin(A,B,C) [C >= 0 ==> && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 1 + C] evalfbbin(A,B,C) = B + C >= B + C = evalfbb1in(A,B,C) [-1 + A + -1*C >= 0 ==> && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] evalfbb1in(A,B,C) = B + C >= B + C = evalfbb3in(A,-1 + B,1 + C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0] ==> evalfreturnin(A,B,C) = B + C >= B + C = evalfstop(A,B,C) * Step 5: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalfstart(A,B,C) -> evalfentryin(A,B,C) True (1,1) 1. evalfentryin(A,B,C) -> evalfbb3in(B,A,0) [A >= 1 && B >= 1] (1,1) 2. evalfbb3in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && 0 >= B] (1,1) 3. evalfbb3in(A,B,C) -> evalfbb4in(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && B >= 1] (?,1) 4. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + D] 5. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && D >= 1] (?,1) 6. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] (1,1) 7. evalfbbin(A,B,C) -> evalfbb1in(A,B,C) [C >= 0 (?,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 1 + C] 8. evalfbbin(A,B,C) -> evalfbb3in(A,B,0) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && C >= A] (A,1) 9. evalfbb1in(A,B,C) -> evalfbb3in(A,-1 + B,1 + C) [-1 + A + -1*C >= 0 (A,1) && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] 10. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0] (1,1) Signature: {(evalfbb1in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfbbin,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{3},2->{10},3->{4,5,6},4->{7,8},5->{7,8},6->{10},7->{9},8->{3},9->{2,3},10->{}] + Applied Processor: KnowledgePropagation + Details: We propagate bounds from predecessors. * Step 6: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalfstart(A,B,C) -> evalfentryin(A,B,C) True (1,1) 1. evalfentryin(A,B,C) -> evalfbb3in(B,A,0) [A >= 1 && B >= 1] (1,1) 2. evalfbb3in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && 0 >= B] (1,1) 3. evalfbb3in(A,B,C) -> evalfbb4in(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0 && B >= 1] (1 + 2*A,1) 4. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 (1 + 2*A,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 0 >= 1 + D] 5. evalfbb4in(A,B,C) -> evalfbbin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && D >= 1] (1 + 2*A,1) 6. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] (1,1) 7. evalfbbin(A,B,C) -> evalfbb1in(A,B,C) [C >= 0 (2 + 4*A,1) && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 1 + C] 8. evalfbbin(A,B,C) -> evalfbb3in(A,B,0) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && C >= A] (A,1) 9. evalfbb1in(A,B,C) -> evalfbb3in(A,-1 + B,1 + C) [-1 + A + -1*C >= 0 (A,1) && C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0] 10. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [C >= 0 && -1 + B + C >= 0 && -1 + A + C >= 0 && -1 + A >= 0] (1,1) Signature: {(evalfbb1in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfbbin,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{3},2->{10},3->{4,5,6},4->{7,8},5->{7,8},6->{10},7->{9},8->{3},9->{2,3},10->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: The problem is already solved. YES(?,O(n^1))