YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalDis1start(A,B,C,D) -> evalDis1entryin(A,B,C,D) True (1,1) 1. evalDis1entryin(A,B,C,D) -> evalDis1bb3in(B,A,D,C) True (?,1) 2. evalDis1bb3in(A,B,C,D) -> evalDis1bbin(A,B,C,D) [A >= 1 + B] (?,1) 3. evalDis1bb3in(A,B,C,D) -> evalDis1returnin(A,B,C,D) [B >= A] (?,1) 4. evalDis1bbin(A,B,C,D) -> evalDis1bb1in(A,B,C,D) [-1 + A + -1*B >= 0 && C >= 1 + D] (?,1) 5. evalDis1bbin(A,B,C,D) -> evalDis1bb2in(A,B,C,D) [-1 + A + -1*B >= 0 && D >= C] (?,1) 6. evalDis1bb1in(A,B,C,D) -> evalDis1bb3in(A,B,C,1 + D) [-1 + C + -1*D >= 0 && -1 + A + -1*B >= 0] (?,1) 7. evalDis1bb2in(A,B,C,D) -> evalDis1bb3in(A,1 + B,C,D) [-1*C + D >= 0 && -1 + A + -1*B >= 0] (?,1) 8. evalDis1returnin(A,B,C,D) -> evalDis1stop(A,B,C,D) [-1*A + B >= 0] (?,1) Signature: {(evalDis1bb1in,4) ;(evalDis1bb2in,4) ;(evalDis1bb3in,4) ;(evalDis1bbin,4) ;(evalDis1entryin,4) ;(evalDis1returnin,4) ;(evalDis1start,4) ;(evalDis1stop,4)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2,3},7->{2,3},8->{}] + 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. evalDis1start(A,B,C,D) -> evalDis1entryin(A,B,C,D) True (1,1) 1. evalDis1entryin(A,B,C,D) -> evalDis1bb3in(B,A,D,C) True (1,1) 2. evalDis1bb3in(A,B,C,D) -> evalDis1bbin(A,B,C,D) [A >= 1 + B] (?,1) 3. evalDis1bb3in(A,B,C,D) -> evalDis1returnin(A,B,C,D) [B >= A] (1,1) 4. evalDis1bbin(A,B,C,D) -> evalDis1bb1in(A,B,C,D) [-1 + A + -1*B >= 0 && C >= 1 + D] (?,1) 5. evalDis1bbin(A,B,C,D) -> evalDis1bb2in(A,B,C,D) [-1 + A + -1*B >= 0 && D >= C] (?,1) 6. evalDis1bb1in(A,B,C,D) -> evalDis1bb3in(A,B,C,1 + D) [-1 + C + -1*D >= 0 && -1 + A + -1*B >= 0] (?,1) 7. evalDis1bb2in(A,B,C,D) -> evalDis1bb3in(A,1 + B,C,D) [-1*C + D >= 0 && -1 + A + -1*B >= 0] (?,1) 8. evalDis1returnin(A,B,C,D) -> evalDis1stop(A,B,C,D) [-1*A + B >= 0] (1,1) Signature: {(evalDis1bb1in,4) ;(evalDis1bb2in,4) ;(evalDis1bb3in,4) ;(evalDis1bbin,4) ;(evalDis1entryin,4) ;(evalDis1returnin,4) ;(evalDis1start,4) ;(evalDis1stop,4)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2,3},7->{2,3},8->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(6,3)] * Step 3: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalDis1start(A,B,C,D) -> evalDis1entryin(A,B,C,D) True (1,1) 1. evalDis1entryin(A,B,C,D) -> evalDis1bb3in(B,A,D,C) True (1,1) 2. evalDis1bb3in(A,B,C,D) -> evalDis1bbin(A,B,C,D) [A >= 1 + B] (?,1) 3. evalDis1bb3in(A,B,C,D) -> evalDis1returnin(A,B,C,D) [B >= A] (1,1) 4. evalDis1bbin(A,B,C,D) -> evalDis1bb1in(A,B,C,D) [-1 + A + -1*B >= 0 && C >= 1 + D] (?,1) 5. evalDis1bbin(A,B,C,D) -> evalDis1bb2in(A,B,C,D) [-1 + A + -1*B >= 0 && D >= C] (?,1) 6. evalDis1bb1in(A,B,C,D) -> evalDis1bb3in(A,B,C,1 + D) [-1 + C + -1*D >= 0 && -1 + A + -1*B >= 0] (?,1) 7. evalDis1bb2in(A,B,C,D) -> evalDis1bb3in(A,1 + B,C,D) [-1*C + D >= 0 && -1 + A + -1*B >= 0] (?,1) 8. evalDis1returnin(A,B,C,D) -> evalDis1stop(A,B,C,D) [-1*A + B >= 0] (1,1) Signature: {(evalDis1bb1in,4) ;(evalDis1bb2in,4) ;(evalDis1bb3in,4) ;(evalDis1bbin,4) ;(evalDis1entryin,4) ;(evalDis1returnin,4) ;(evalDis1start,4) ;(evalDis1stop,4)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2},7->{2,3},8->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(evalDis1bb1in) = x1 + -1*x2 p(evalDis1bb2in) = x1 + -1*x2 p(evalDis1bb3in) = x1 + -1*x2 p(evalDis1bbin) = x1 + -1*x2 p(evalDis1entryin) = -1*x1 + x2 p(evalDis1returnin) = x1 + -1*x2 p(evalDis1start) = -1*x1 + x2 p(evalDis1stop) = x1 + -1*x2 Following rules are strictly oriented: [-1*C + D >= 0 && -1 + A + -1*B >= 0] ==> evalDis1bb2in(A,B,C,D) = A + -1*B > -1 + A + -1*B = evalDis1bb3in(A,1 + B,C,D) Following rules are weakly oriented: True ==> evalDis1start(A,B,C,D) = -1*A + B >= -1*A + B = evalDis1entryin(A,B,C,D) True ==> evalDis1entryin(A,B,C,D) = -1*A + B >= -1*A + B = evalDis1bb3in(B,A,D,C) [A >= 1 + B] ==> evalDis1bb3in(A,B,C,D) = A + -1*B >= A + -1*B = evalDis1bbin(A,B,C,D) [B >= A] ==> evalDis1bb3in(A,B,C,D) = A + -1*B >= A + -1*B = evalDis1returnin(A,B,C,D) [-1 + A + -1*B >= 0 && C >= 1 + D] ==> evalDis1bbin(A,B,C,D) = A + -1*B >= A + -1*B = evalDis1bb1in(A,B,C,D) [-1 + A + -1*B >= 0 && D >= C] ==> evalDis1bbin(A,B,C,D) = A + -1*B >= A + -1*B = evalDis1bb2in(A,B,C,D) [-1 + C + -1*D >= 0 && -1 + A + -1*B >= 0] ==> evalDis1bb1in(A,B,C,D) = A + -1*B >= A + -1*B = evalDis1bb3in(A,B,C,1 + D) [-1*A + B >= 0] ==> evalDis1returnin(A,B,C,D) = A + -1*B >= A + -1*B = evalDis1stop(A,B,C,D) * Step 4: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalDis1start(A,B,C,D) -> evalDis1entryin(A,B,C,D) True (1,1) 1. evalDis1entryin(A,B,C,D) -> evalDis1bb3in(B,A,D,C) True (1,1) 2. evalDis1bb3in(A,B,C,D) -> evalDis1bbin(A,B,C,D) [A >= 1 + B] (?,1) 3. evalDis1bb3in(A,B,C,D) -> evalDis1returnin(A,B,C,D) [B >= A] (1,1) 4. evalDis1bbin(A,B,C,D) -> evalDis1bb1in(A,B,C,D) [-1 + A + -1*B >= 0 && C >= 1 + D] (?,1) 5. evalDis1bbin(A,B,C,D) -> evalDis1bb2in(A,B,C,D) [-1 + A + -1*B >= 0 && D >= C] (?,1) 6. evalDis1bb1in(A,B,C,D) -> evalDis1bb3in(A,B,C,1 + D) [-1 + C + -1*D >= 0 && -1 + A + -1*B >= 0] (?,1) 7. evalDis1bb2in(A,B,C,D) -> evalDis1bb3in(A,1 + B,C,D) [-1*C + D >= 0 && -1 + A + -1*B >= 0] (A + B,1) 8. evalDis1returnin(A,B,C,D) -> evalDis1stop(A,B,C,D) [-1*A + B >= 0] (1,1) Signature: {(evalDis1bb1in,4) ;(evalDis1bb2in,4) ;(evalDis1bb3in,4) ;(evalDis1bbin,4) ;(evalDis1entryin,4) ;(evalDis1returnin,4) ;(evalDis1start,4) ;(evalDis1stop,4)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2},7->{2,3},8->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(evalDis1bb1in) = x3 + -1*x4 p(evalDis1bb2in) = x3 + -1*x4 p(evalDis1bb3in) = x3 + -1*x4 p(evalDis1bbin) = x3 + -1*x4 p(evalDis1entryin) = -1*x3 + x4 p(evalDis1returnin) = x3 + -1*x4 p(evalDis1start) = -1*x3 + x4 p(evalDis1stop) = x3 + -1*x4 Following rules are strictly oriented: [-1 + C + -1*D >= 0 && -1 + A + -1*B >= 0] ==> evalDis1bb1in(A,B,C,D) = C + -1*D > -1 + C + -1*D = evalDis1bb3in(A,B,C,1 + D) Following rules are weakly oriented: True ==> evalDis1start(A,B,C,D) = -1*C + D >= -1*C + D = evalDis1entryin(A,B,C,D) True ==> evalDis1entryin(A,B,C,D) = -1*C + D >= -1*C + D = evalDis1bb3in(B,A,D,C) [A >= 1 + B] ==> evalDis1bb3in(A,B,C,D) = C + -1*D >= C + -1*D = evalDis1bbin(A,B,C,D) [B >= A] ==> evalDis1bb3in(A,B,C,D) = C + -1*D >= C + -1*D = evalDis1returnin(A,B,C,D) [-1 + A + -1*B >= 0 && C >= 1 + D] ==> evalDis1bbin(A,B,C,D) = C + -1*D >= C + -1*D = evalDis1bb1in(A,B,C,D) [-1 + A + -1*B >= 0 && D >= C] ==> evalDis1bbin(A,B,C,D) = C + -1*D >= C + -1*D = evalDis1bb2in(A,B,C,D) [-1*C + D >= 0 && -1 + A + -1*B >= 0] ==> evalDis1bb2in(A,B,C,D) = C + -1*D >= C + -1*D = evalDis1bb3in(A,1 + B,C,D) [-1*A + B >= 0] ==> evalDis1returnin(A,B,C,D) = C + -1*D >= C + -1*D = evalDis1stop(A,B,C,D) * Step 5: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalDis1start(A,B,C,D) -> evalDis1entryin(A,B,C,D) True (1,1) 1. evalDis1entryin(A,B,C,D) -> evalDis1bb3in(B,A,D,C) True (1,1) 2. evalDis1bb3in(A,B,C,D) -> evalDis1bbin(A,B,C,D) [A >= 1 + B] (?,1) 3. evalDis1bb3in(A,B,C,D) -> evalDis1returnin(A,B,C,D) [B >= A] (1,1) 4. evalDis1bbin(A,B,C,D) -> evalDis1bb1in(A,B,C,D) [-1 + A + -1*B >= 0 && C >= 1 + D] (?,1) 5. evalDis1bbin(A,B,C,D) -> evalDis1bb2in(A,B,C,D) [-1 + A + -1*B >= 0 && D >= C] (?,1) 6. evalDis1bb1in(A,B,C,D) -> evalDis1bb3in(A,B,C,1 + D) [-1 + C + -1*D >= 0 && -1 + A + -1*B >= 0] (C + D,1) 7. evalDis1bb2in(A,B,C,D) -> evalDis1bb3in(A,1 + B,C,D) [-1*C + D >= 0 && -1 + A + -1*B >= 0] (A + B,1) 8. evalDis1returnin(A,B,C,D) -> evalDis1stop(A,B,C,D) [-1*A + B >= 0] (1,1) Signature: {(evalDis1bb1in,4) ;(evalDis1bb2in,4) ;(evalDis1bb3in,4) ;(evalDis1bbin,4) ;(evalDis1entryin,4) ;(evalDis1returnin,4) ;(evalDis1start,4) ;(evalDis1stop,4)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2},7->{2,3},8->{}] + Applied Processor: KnowledgePropagation + Details: We propagate bounds from predecessors. * Step 6: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalDis1start(A,B,C,D) -> evalDis1entryin(A,B,C,D) True (1,1) 1. evalDis1entryin(A,B,C,D) -> evalDis1bb3in(B,A,D,C) True (1,1) 2. evalDis1bb3in(A,B,C,D) -> evalDis1bbin(A,B,C,D) [A >= 1 + B] (1 + A + B + C + D,1) 3. evalDis1bb3in(A,B,C,D) -> evalDis1returnin(A,B,C,D) [B >= A] (1,1) 4. evalDis1bbin(A,B,C,D) -> evalDis1bb1in(A,B,C,D) [-1 + A + -1*B >= 0 && C >= 1 + D] (1 + A + B + C + D,1) 5. evalDis1bbin(A,B,C,D) -> evalDis1bb2in(A,B,C,D) [-1 + A + -1*B >= 0 && D >= C] (1 + A + B + C + D,1) 6. evalDis1bb1in(A,B,C,D) -> evalDis1bb3in(A,B,C,1 + D) [-1 + C + -1*D >= 0 && -1 + A + -1*B >= 0] (C + D,1) 7. evalDis1bb2in(A,B,C,D) -> evalDis1bb3in(A,1 + B,C,D) [-1*C + D >= 0 && -1 + A + -1*B >= 0] (A + B,1) 8. evalDis1returnin(A,B,C,D) -> evalDis1stop(A,B,C,D) [-1*A + B >= 0] (1,1) Signature: {(evalDis1bb1in,4) ;(evalDis1bb2in,4) ;(evalDis1bb3in,4) ;(evalDis1bbin,4) ;(evalDis1entryin,4) ;(evalDis1returnin,4) ;(evalDis1start,4) ;(evalDis1stop,4)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2},7->{2,3},8->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: The problem is already solved. YES(?,O(n^1))