YES(?,O(n^1)) * Step 1: 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) -> evalfbb4in(B,A,C) True (?,1) 2. evalfbb4in(A,B,C) -> evalfbb2in(A,B,A) [B >= 1] (?,1) 3. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [0 >= B] (?,1) 4. evalfbb2in(A,B,C) -> evalfbb1in(A,B,C) [A + -1*C >= 0 && -1 + B >= 0 && C >= A] (?,1) 5. evalfbb2in(A,B,C) -> evalfbb3in(A,B,C) [A + -1*C >= 0 && -1 + B >= 0 && A >= 1 + C] (?,1) 6. evalfbb1in(A,B,C) -> evalfbb2in(A,B,-1 + C) [A + -1*C >= 0 && -1*A + C >= 0 && -1 + B >= 0] (?,1) 7. evalfbb3in(A,B,C) -> evalfbb4in(A,-1 + B,C) [-1 + A + -1*C >= 0 && -1 + B >= 0] (?,1) 8. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [-1*B >= 0] (?,1) Signature: {(evalfbb1in,3) ;(evalfbb2in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{4,5},7->{2,3},8->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(2,5),(6,4)] * Step 2: 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) -> evalfbb4in(B,A,C) True (?,1) 2. evalfbb4in(A,B,C) -> evalfbb2in(A,B,A) [B >= 1] (?,1) 3. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [0 >= B] (?,1) 4. evalfbb2in(A,B,C) -> evalfbb1in(A,B,C) [A + -1*C >= 0 && -1 + B >= 0 && C >= A] (?,1) 5. evalfbb2in(A,B,C) -> evalfbb3in(A,B,C) [A + -1*C >= 0 && -1 + B >= 0 && A >= 1 + C] (?,1) 6. evalfbb1in(A,B,C) -> evalfbb2in(A,B,-1 + C) [A + -1*C >= 0 && -1*A + C >= 0 && -1 + B >= 0] (?,1) 7. evalfbb3in(A,B,C) -> evalfbb4in(A,-1 + B,C) [-1 + A + -1*C >= 0 && -1 + B >= 0] (?,1) 8. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [-1*B >= 0] (?,1) Signature: {(evalfbb1in,3) ;(evalfbb2in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{2,3},2->{4},3->{8},4->{6},5->{7},6->{5},7->{2,3},8->{}] + 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. evalfstart(A,B,C) -> evalfentryin(A,B,C) True (1,1) 1. evalfentryin(A,B,C) -> evalfbb4in(B,A,C) True (1,1) 2. evalfbb4in(A,B,C) -> evalfbb2in(A,B,A) [B >= 1] (?,1) 3. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [0 >= B] (1,1) 4. evalfbb2in(A,B,C) -> evalfbb1in(A,B,C) [A + -1*C >= 0 && -1 + B >= 0 && C >= A] (?,1) 5. evalfbb2in(A,B,C) -> evalfbb3in(A,B,C) [A + -1*C >= 0 && -1 + B >= 0 && A >= 1 + C] (?,1) 6. evalfbb1in(A,B,C) -> evalfbb2in(A,B,-1 + C) [A + -1*C >= 0 && -1*A + C >= 0 && -1 + B >= 0] (?,1) 7. evalfbb3in(A,B,C) -> evalfbb4in(A,-1 + B,C) [-1 + A + -1*C >= 0 && -1 + B >= 0] (?,1) 8. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [-1*B >= 0] (1,1) Signature: {(evalfbb1in,3) ;(evalfbb2in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{2,3},2->{4},3->{8},4->{6},5->{7},6->{5},7->{2,3},8->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(evalfbb1in) = x2 p(evalfbb2in) = x2 p(evalfbb3in) = x2 p(evalfbb4in) = x2 p(evalfentryin) = x1 p(evalfreturnin) = x2 p(evalfstart) = x1 p(evalfstop) = x2 Following rules are strictly oriented: [-1 + A + -1*C >= 0 && -1 + B >= 0] ==> evalfbb3in(A,B,C) = B > -1 + B = evalfbb4in(A,-1 + B,C) Following rules are weakly oriented: True ==> evalfstart(A,B,C) = A >= A = evalfentryin(A,B,C) True ==> evalfentryin(A,B,C) = A >= A = evalfbb4in(B,A,C) [B >= 1] ==> evalfbb4in(A,B,C) = B >= B = evalfbb2in(A,B,A) [0 >= B] ==> evalfbb4in(A,B,C) = B >= B = evalfreturnin(A,B,C) [A + -1*C >= 0 && -1 + B >= 0 && C >= A] ==> evalfbb2in(A,B,C) = B >= B = evalfbb1in(A,B,C) [A + -1*C >= 0 && -1 + B >= 0 && A >= 1 + C] ==> evalfbb2in(A,B,C) = B >= B = evalfbb3in(A,B,C) [A + -1*C >= 0 && -1*A + C >= 0 && -1 + B >= 0] ==> evalfbb1in(A,B,C) = B >= B = evalfbb2in(A,B,-1 + C) [-1*B >= 0] ==> evalfreturnin(A,B,C) = B >= B = evalfstop(A,B,C) * Step 4: 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) -> evalfbb4in(B,A,C) True (1,1) 2. evalfbb4in(A,B,C) -> evalfbb2in(A,B,A) [B >= 1] (?,1) 3. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [0 >= B] (1,1) 4. evalfbb2in(A,B,C) -> evalfbb1in(A,B,C) [A + -1*C >= 0 && -1 + B >= 0 && C >= A] (?,1) 5. evalfbb2in(A,B,C) -> evalfbb3in(A,B,C) [A + -1*C >= 0 && -1 + B >= 0 && A >= 1 + C] (?,1) 6. evalfbb1in(A,B,C) -> evalfbb2in(A,B,-1 + C) [A + -1*C >= 0 && -1*A + C >= 0 && -1 + B >= 0] (?,1) 7. evalfbb3in(A,B,C) -> evalfbb4in(A,-1 + B,C) [-1 + A + -1*C >= 0 && -1 + B >= 0] (A,1) 8. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [-1*B >= 0] (1,1) Signature: {(evalfbb1in,3) ;(evalfbb2in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{2,3},2->{4},3->{8},4->{6},5->{7},6->{5},7->{2,3},8->{}] + Applied Processor: KnowledgePropagation + Details: We propagate bounds from predecessors. * Step 5: 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) -> evalfbb4in(B,A,C) True (1,1) 2. evalfbb4in(A,B,C) -> evalfbb2in(A,B,A) [B >= 1] (1 + A,1) 3. evalfbb4in(A,B,C) -> evalfreturnin(A,B,C) [0 >= B] (1,1) 4. evalfbb2in(A,B,C) -> evalfbb1in(A,B,C) [A + -1*C >= 0 && -1 + B >= 0 && C >= A] (1 + A,1) 5. evalfbb2in(A,B,C) -> evalfbb3in(A,B,C) [A + -1*C >= 0 && -1 + B >= 0 && A >= 1 + C] (1 + A,1) 6. evalfbb1in(A,B,C) -> evalfbb2in(A,B,-1 + C) [A + -1*C >= 0 && -1*A + C >= 0 && -1 + B >= 0] (1 + A,1) 7. evalfbb3in(A,B,C) -> evalfbb4in(A,-1 + B,C) [-1 + A + -1*C >= 0 && -1 + B >= 0] (A,1) 8. evalfreturnin(A,B,C) -> evalfstop(A,B,C) [-1*B >= 0] (1,1) Signature: {(evalfbb1in,3) ;(evalfbb2in,3) ;(evalfbb3in,3) ;(evalfbb4in,3) ;(evalfentryin,3) ;(evalfreturnin,3) ;(evalfstart,3) ;(evalfstop,3)} Flow Graph: [0->{1},1->{2,3},2->{4},3->{8},4->{6},5->{7},6->{5},7->{2,3},8->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: The problem is already solved. YES(?,O(n^1))