YES(?,O(n^1)) * Step 1: UnsatPaths WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. eval1(A,B) -> eval2(A,B) [A >= 1 && B >= 1 && A >= 1 + B] (?,1) 1. eval1(A,B) -> eval3(A,B) [A >= 1 && B >= 1 && B >= A] (?,1) 2. eval2(A,B) -> eval2(-1 + A,B) [-1 + B >= 0 && A >= 1] (?,1) 3. eval2(A,B) -> eval1(A,B) [-1 + B >= 0 && 0 >= A] (?,1) 4. eval3(A,B) -> eval3(A,-1 + B) [-1 + A >= 0 && B >= 1] (?,1) 5. eval3(A,B) -> eval1(A,B) [-1 + A >= 0 && 0 >= B] (?,1) 6. start(A,B) -> eval1(A,B) True (1,1) Signature: {(eval1,2);(eval2,2);(eval3,2);(start,2)} Flow Graph: [0->{2,3},1->{4,5},2->{2,3},3->{0,1},4->{4,5},5->{0,1},6->{0,1}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(0,3),(1,5),(3,0),(3,1),(5,0),(5,1)] * Step 2: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. eval1(A,B) -> eval2(A,B) [A >= 1 && B >= 1 && A >= 1 + B] (?,1) 1. eval1(A,B) -> eval3(A,B) [A >= 1 && B >= 1 && B >= A] (?,1) 2. eval2(A,B) -> eval2(-1 + A,B) [-1 + B >= 0 && A >= 1] (?,1) 3. eval2(A,B) -> eval1(A,B) [-1 + B >= 0 && 0 >= A] (?,1) 4. eval3(A,B) -> eval3(A,-1 + B) [-1 + A >= 0 && B >= 1] (?,1) 5. eval3(A,B) -> eval1(A,B) [-1 + A >= 0 && 0 >= B] (?,1) 6. start(A,B) -> eval1(A,B) True (1,1) Signature: {(eval1,2);(eval2,2);(eval3,2);(start,2)} Flow Graph: [0->{2},1->{4},2->{2,3},3->{},4->{4,5},5->{},6->{0,1}] + 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. eval1(A,B) -> eval2(A,B) [A >= 1 && B >= 1 && A >= 1 + B] (1,1) 1. eval1(A,B) -> eval3(A,B) [A >= 1 && B >= 1 && B >= A] (1,1) 2. eval2(A,B) -> eval2(-1 + A,B) [-1 + B >= 0 && A >= 1] (?,1) 3. eval2(A,B) -> eval1(A,B) [-1 + B >= 0 && 0 >= A] (1,1) 4. eval3(A,B) -> eval3(A,-1 + B) [-1 + A >= 0 && B >= 1] (?,1) 5. eval3(A,B) -> eval1(A,B) [-1 + A >= 0 && 0 >= B] (1,1) 6. start(A,B) -> eval1(A,B) True (1,1) Signature: {(eval1,2);(eval2,2);(eval3,2);(start,2)} Flow Graph: [0->{2},1->{4},2->{2,3},3->{},4->{4,5},5->{},6->{0,1}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(eval1) = x2 p(eval2) = x2 p(eval3) = x2 p(start) = x2 Following rules are strictly oriented: [-1 + A >= 0 && B >= 1] ==> eval3(A,B) = B > -1 + B = eval3(A,-1 + B) Following rules are weakly oriented: [A >= 1 && B >= 1 && A >= 1 + B] ==> eval1(A,B) = B >= B = eval2(A,B) [A >= 1 && B >= 1 && B >= A] ==> eval1(A,B) = B >= B = eval3(A,B) [-1 + B >= 0 && A >= 1] ==> eval2(A,B) = B >= B = eval2(-1 + A,B) [-1 + B >= 0 && 0 >= A] ==> eval2(A,B) = B >= B = eval1(A,B) [-1 + A >= 0 && 0 >= B] ==> eval3(A,B) = B >= B = eval1(A,B) True ==> start(A,B) = B >= B = eval1(A,B) * Step 4: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. eval1(A,B) -> eval2(A,B) [A >= 1 && B >= 1 && A >= 1 + B] (1,1) 1. eval1(A,B) -> eval3(A,B) [A >= 1 && B >= 1 && B >= A] (1,1) 2. eval2(A,B) -> eval2(-1 + A,B) [-1 + B >= 0 && A >= 1] (?,1) 3. eval2(A,B) -> eval1(A,B) [-1 + B >= 0 && 0 >= A] (1,1) 4. eval3(A,B) -> eval3(A,-1 + B) [-1 + A >= 0 && B >= 1] (B,1) 5. eval3(A,B) -> eval1(A,B) [-1 + A >= 0 && 0 >= B] (1,1) 6. start(A,B) -> eval1(A,B) True (1,1) Signature: {(eval1,2);(eval2,2);(eval3,2);(start,2)} Flow Graph: [0->{2},1->{4},2->{2,3},3->{},4->{4,5},5->{},6->{0,1}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(eval1) = x1 p(eval2) = x1 p(eval3) = x1 p(start) = x1 Following rules are strictly oriented: [-1 + B >= 0 && A >= 1] ==> eval2(A,B) = A > -1 + A = eval2(-1 + A,B) Following rules are weakly oriented: [A >= 1 && B >= 1 && A >= 1 + B] ==> eval1(A,B) = A >= A = eval2(A,B) [A >= 1 && B >= 1 && B >= A] ==> eval1(A,B) = A >= A = eval3(A,B) [-1 + B >= 0 && 0 >= A] ==> eval2(A,B) = A >= A = eval1(A,B) [-1 + A >= 0 && B >= 1] ==> eval3(A,B) = A >= A = eval3(A,-1 + B) [-1 + A >= 0 && 0 >= B] ==> eval3(A,B) = A >= A = eval1(A,B) True ==> start(A,B) = A >= A = eval1(A,B) * Step 5: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. eval1(A,B) -> eval2(A,B) [A >= 1 && B >= 1 && A >= 1 + B] (1,1) 1. eval1(A,B) -> eval3(A,B) [A >= 1 && B >= 1 && B >= A] (1,1) 2. eval2(A,B) -> eval2(-1 + A,B) [-1 + B >= 0 && A >= 1] (A,1) 3. eval2(A,B) -> eval1(A,B) [-1 + B >= 0 && 0 >= A] (1,1) 4. eval3(A,B) -> eval3(A,-1 + B) [-1 + A >= 0 && B >= 1] (B,1) 5. eval3(A,B) -> eval1(A,B) [-1 + A >= 0 && 0 >= B] (1,1) 6. start(A,B) -> eval1(A,B) True (1,1) Signature: {(eval1,2);(eval2,2);(eval3,2);(start,2)} Flow Graph: [0->{2},1->{4},2->{2,3},3->{},4->{4,5},5->{},6->{0,1}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))