YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalDis2start(A,B,C) -> evalDis2entryin(A,B,C) True (1,1) 1. evalDis2entryin(A,B,C) -> evalDis2bb3in(B,C,A) True (?,1) 2. evalDis2bb3in(A,B,C) -> evalDis2bbin(A,B,C) [A >= 1 + C] (?,1) 3. evalDis2bb3in(A,B,C) -> evalDis2returnin(A,B,C) [C >= A] (?,1) 4. evalDis2bbin(A,B,C) -> evalDis2bb1in(A,B,C) [-1 + A + -1*C >= 0 && B >= 1 + C] (?,1) 5. evalDis2bbin(A,B,C) -> evalDis2bb2in(A,B,C) [-1 + A + -1*C >= 0 && C >= B] (?,1) 6. evalDis2bb1in(A,B,C) -> evalDis2bb3in(A,B,1 + C) [-1 + B + -1*C >= 0 && -1 + A + -1*C >= 0] (?,1) 7. evalDis2bb2in(A,B,C) -> evalDis2bb3in(A,1 + B,C) [-1 + A + -1*C >= 0 && -1*B + C >= 0 && -1 + A + -1*B >= 0] (?,1) 8. evalDis2returnin(A,B,C) -> evalDis2stop(A,B,C) [-1*A + C >= 0] (?,1) Signature: {(evalDis2bb1in,3) ;(evalDis2bb2in,3) ;(evalDis2bb3in,3) ;(evalDis2bbin,3) ;(evalDis2entryin,3) ;(evalDis2returnin,3) ;(evalDis2start,3) ;(evalDis2stop,3)} 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. evalDis2start(A,B,C) -> evalDis2entryin(A,B,C) True (1,1) 1. evalDis2entryin(A,B,C) -> evalDis2bb3in(B,C,A) True (1,1) 2. evalDis2bb3in(A,B,C) -> evalDis2bbin(A,B,C) [A >= 1 + C] (?,1) 3. evalDis2bb3in(A,B,C) -> evalDis2returnin(A,B,C) [C >= A] (1,1) 4. evalDis2bbin(A,B,C) -> evalDis2bb1in(A,B,C) [-1 + A + -1*C >= 0 && B >= 1 + C] (?,1) 5. evalDis2bbin(A,B,C) -> evalDis2bb2in(A,B,C) [-1 + A + -1*C >= 0 && C >= B] (?,1) 6. evalDis2bb1in(A,B,C) -> evalDis2bb3in(A,B,1 + C) [-1 + B + -1*C >= 0 && -1 + A + -1*C >= 0] (?,1) 7. evalDis2bb2in(A,B,C) -> evalDis2bb3in(A,1 + B,C) [-1 + A + -1*C >= 0 && -1*B + C >= 0 && -1 + A + -1*B >= 0] (?,1) 8. evalDis2returnin(A,B,C) -> evalDis2stop(A,B,C) [-1*A + C >= 0] (1,1) Signature: {(evalDis2bb1in,3) ;(evalDis2bb2in,3) ;(evalDis2bb3in,3) ;(evalDis2bbin,3) ;(evalDis2entryin,3) ;(evalDis2returnin,3) ;(evalDis2start,3) ;(evalDis2stop,3)} 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: [(7,3)] * Step 3: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalDis2start(A,B,C) -> evalDis2entryin(A,B,C) True (1,1) 1. evalDis2entryin(A,B,C) -> evalDis2bb3in(B,C,A) True (1,1) 2. evalDis2bb3in(A,B,C) -> evalDis2bbin(A,B,C) [A >= 1 + C] (?,1) 3. evalDis2bb3in(A,B,C) -> evalDis2returnin(A,B,C) [C >= A] (1,1) 4. evalDis2bbin(A,B,C) -> evalDis2bb1in(A,B,C) [-1 + A + -1*C >= 0 && B >= 1 + C] (?,1) 5. evalDis2bbin(A,B,C) -> evalDis2bb2in(A,B,C) [-1 + A + -1*C >= 0 && C >= B] (?,1) 6. evalDis2bb1in(A,B,C) -> evalDis2bb3in(A,B,1 + C) [-1 + B + -1*C >= 0 && -1 + A + -1*C >= 0] (?,1) 7. evalDis2bb2in(A,B,C) -> evalDis2bb3in(A,1 + B,C) [-1 + A + -1*C >= 0 && -1*B + C >= 0 && -1 + A + -1*B >= 0] (?,1) 8. evalDis2returnin(A,B,C) -> evalDis2stop(A,B,C) [-1*A + C >= 0] (1,1) Signature: {(evalDis2bb1in,3) ;(evalDis2bb2in,3) ;(evalDis2bb3in,3) ;(evalDis2bbin,3) ;(evalDis2entryin,3) ;(evalDis2returnin,3) ;(evalDis2start,3) ;(evalDis2stop,3)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2,3},7->{2},8->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(evalDis2bb1in) = x1 + -1*x2 p(evalDis2bb2in) = x1 + -1*x2 p(evalDis2bb3in) = x1 + -1*x2 p(evalDis2bbin) = x1 + -1*x2 p(evalDis2entryin) = x2 + -1*x3 p(evalDis2returnin) = x1 + -1*x2 p(evalDis2start) = x2 + -1*x3 p(evalDis2stop) = x1 + -1*x2 Following rules are strictly oriented: [-1 + A + -1*C >= 0 && -1*B + C >= 0 && -1 + A + -1*B >= 0] ==> evalDis2bb2in(A,B,C) = A + -1*B > -1 + A + -1*B = evalDis2bb3in(A,1 + B,C) Following rules are weakly oriented: True ==> evalDis2start(A,B,C) = B + -1*C >= B + -1*C = evalDis2entryin(A,B,C) True ==> evalDis2entryin(A,B,C) = B + -1*C >= B + -1*C = evalDis2bb3in(B,C,A) [A >= 1 + C] ==> evalDis2bb3in(A,B,C) = A + -1*B >= A + -1*B = evalDis2bbin(A,B,C) [C >= A] ==> evalDis2bb3in(A,B,C) = A + -1*B >= A + -1*B = evalDis2returnin(A,B,C) [-1 + A + -1*C >= 0 && B >= 1 + C] ==> evalDis2bbin(A,B,C) = A + -1*B >= A + -1*B = evalDis2bb1in(A,B,C) [-1 + A + -1*C >= 0 && C >= B] ==> evalDis2bbin(A,B,C) = A + -1*B >= A + -1*B = evalDis2bb2in(A,B,C) [-1 + B + -1*C >= 0 && -1 + A + -1*C >= 0] ==> evalDis2bb1in(A,B,C) = A + -1*B >= A + -1*B = evalDis2bb3in(A,B,1 + C) [-1*A + C >= 0] ==> evalDis2returnin(A,B,C) = A + -1*B >= A + -1*B = evalDis2stop(A,B,C) * Step 4: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalDis2start(A,B,C) -> evalDis2entryin(A,B,C) True (1,1) 1. evalDis2entryin(A,B,C) -> evalDis2bb3in(B,C,A) True (1,1) 2. evalDis2bb3in(A,B,C) -> evalDis2bbin(A,B,C) [A >= 1 + C] (?,1) 3. evalDis2bb3in(A,B,C) -> evalDis2returnin(A,B,C) [C >= A] (1,1) 4. evalDis2bbin(A,B,C) -> evalDis2bb1in(A,B,C) [-1 + A + -1*C >= 0 && B >= 1 + C] (?,1) 5. evalDis2bbin(A,B,C) -> evalDis2bb2in(A,B,C) [-1 + A + -1*C >= 0 && C >= B] (?,1) 6. evalDis2bb1in(A,B,C) -> evalDis2bb3in(A,B,1 + C) [-1 + B + -1*C >= 0 && -1 + A + -1*C >= 0] (?,1) 7. evalDis2bb2in(A,B,C) -> evalDis2bb3in(A,1 + B,C) [-1 + A + -1*C >= 0 && -1*B + C >= 0 && -1 + A + -1*B >= 0] (B + C,1) 8. evalDis2returnin(A,B,C) -> evalDis2stop(A,B,C) [-1*A + C >= 0] (1,1) Signature: {(evalDis2bb1in,3) ;(evalDis2bb2in,3) ;(evalDis2bb3in,3) ;(evalDis2bbin,3) ;(evalDis2entryin,3) ;(evalDis2returnin,3) ;(evalDis2start,3) ;(evalDis2stop,3)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2,3},7->{2},8->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(evalDis2bb1in) = x1 + -1*x3 p(evalDis2bb2in) = x1 + -1*x3 p(evalDis2bb3in) = x1 + -1*x3 p(evalDis2bbin) = x1 + -1*x3 p(evalDis2entryin) = -1*x1 + x2 p(evalDis2returnin) = x1 + -1*x3 p(evalDis2start) = -1*x1 + x2 p(evalDis2stop) = x1 + -1*x3 Following rules are strictly oriented: [-1 + B + -1*C >= 0 && -1 + A + -1*C >= 0] ==> evalDis2bb1in(A,B,C) = A + -1*C > -1 + A + -1*C = evalDis2bb3in(A,B,1 + C) Following rules are weakly oriented: True ==> evalDis2start(A,B,C) = -1*A + B >= -1*A + B = evalDis2entryin(A,B,C) True ==> evalDis2entryin(A,B,C) = -1*A + B >= -1*A + B = evalDis2bb3in(B,C,A) [A >= 1 + C] ==> evalDis2bb3in(A,B,C) = A + -1*C >= A + -1*C = evalDis2bbin(A,B,C) [C >= A] ==> evalDis2bb3in(A,B,C) = A + -1*C >= A + -1*C = evalDis2returnin(A,B,C) [-1 + A + -1*C >= 0 && B >= 1 + C] ==> evalDis2bbin(A,B,C) = A + -1*C >= A + -1*C = evalDis2bb1in(A,B,C) [-1 + A + -1*C >= 0 && C >= B] ==> evalDis2bbin(A,B,C) = A + -1*C >= A + -1*C = evalDis2bb2in(A,B,C) [-1 + A + -1*C >= 0 && -1*B + C >= 0 && -1 + A + -1*B >= 0] ==> evalDis2bb2in(A,B,C) = A + -1*C >= A + -1*C = evalDis2bb3in(A,1 + B,C) [-1*A + C >= 0] ==> evalDis2returnin(A,B,C) = A + -1*C >= A + -1*C = evalDis2stop(A,B,C) * Step 5: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalDis2start(A,B,C) -> evalDis2entryin(A,B,C) True (1,1) 1. evalDis2entryin(A,B,C) -> evalDis2bb3in(B,C,A) True (1,1) 2. evalDis2bb3in(A,B,C) -> evalDis2bbin(A,B,C) [A >= 1 + C] (?,1) 3. evalDis2bb3in(A,B,C) -> evalDis2returnin(A,B,C) [C >= A] (1,1) 4. evalDis2bbin(A,B,C) -> evalDis2bb1in(A,B,C) [-1 + A + -1*C >= 0 && B >= 1 + C] (?,1) 5. evalDis2bbin(A,B,C) -> evalDis2bb2in(A,B,C) [-1 + A + -1*C >= 0 && C >= B] (?,1) 6. evalDis2bb1in(A,B,C) -> evalDis2bb3in(A,B,1 + C) [-1 + B + -1*C >= 0 && -1 + A + -1*C >= 0] (A + B,1) 7. evalDis2bb2in(A,B,C) -> evalDis2bb3in(A,1 + B,C) [-1 + A + -1*C >= 0 && -1*B + C >= 0 && -1 + A + -1*B >= 0] (B + C,1) 8. evalDis2returnin(A,B,C) -> evalDis2stop(A,B,C) [-1*A + C >= 0] (1,1) Signature: {(evalDis2bb1in,3) ;(evalDis2bb2in,3) ;(evalDis2bb3in,3) ;(evalDis2bbin,3) ;(evalDis2entryin,3) ;(evalDis2returnin,3) ;(evalDis2start,3) ;(evalDis2stop,3)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2,3},7->{2},8->{}] + Applied Processor: KnowledgePropagation + Details: We propagate bounds from predecessors. * Step 6: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. evalDis2start(A,B,C) -> evalDis2entryin(A,B,C) True (1,1) 1. evalDis2entryin(A,B,C) -> evalDis2bb3in(B,C,A) True (1,1) 2. evalDis2bb3in(A,B,C) -> evalDis2bbin(A,B,C) [A >= 1 + C] (1 + A + 2*B + C,1) 3. evalDis2bb3in(A,B,C) -> evalDis2returnin(A,B,C) [C >= A] (1,1) 4. evalDis2bbin(A,B,C) -> evalDis2bb1in(A,B,C) [-1 + A + -1*C >= 0 && B >= 1 + C] (1 + A + 2*B + C,1) 5. evalDis2bbin(A,B,C) -> evalDis2bb2in(A,B,C) [-1 + A + -1*C >= 0 && C >= B] (1 + A + 2*B + C,1) 6. evalDis2bb1in(A,B,C) -> evalDis2bb3in(A,B,1 + C) [-1 + B + -1*C >= 0 && -1 + A + -1*C >= 0] (A + B,1) 7. evalDis2bb2in(A,B,C) -> evalDis2bb3in(A,1 + B,C) [-1 + A + -1*C >= 0 && -1*B + C >= 0 && -1 + A + -1*B >= 0] (B + C,1) 8. evalDis2returnin(A,B,C) -> evalDis2stop(A,B,C) [-1*A + C >= 0] (1,1) Signature: {(evalDis2bb1in,3) ;(evalDis2bb2in,3) ;(evalDis2bb3in,3) ;(evalDis2bbin,3) ;(evalDis2entryin,3) ;(evalDis2returnin,3) ;(evalDis2start,3) ;(evalDis2stop,3)} Flow Graph: [0->{1},1->{2,3},2->{4,5},3->{8},4->{6},5->{7},6->{2,3},7->{2},8->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: The problem is already solved. YES(?,O(n^1))