YES(?,O(n^1)) * Step 1: UnsatPaths WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f2(A,B,C,D,E) -> f1(A,F,C,D,E) [1 >= A] (1,1) 1. f2(A,B,C,D,E) -> f300(A,B,C,D,E) [A >= 2 && C >= 2] (1,1) 2. f300(A,B,C,D,E) -> f1(A,F,C,D,E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0] (?,1) 3. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D] (?,1) 4. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] (?,1) Signature: {(f1,5);(f2,5);(f300,5)} Flow Graph: [0->{},1->{2,3,4},2->{},3->{2,3,4},4->{2,3,4}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(3,4),(4,3)] * Step 2: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f2(A,B,C,D,E) -> f1(A,F,C,D,E) [1 >= A] (1,1) 1. f2(A,B,C,D,E) -> f300(A,B,C,D,E) [A >= 2 && C >= 2] (1,1) 2. f300(A,B,C,D,E) -> f1(A,F,C,D,E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0] (?,1) 3. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D] (?,1) 4. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] (?,1) Signature: {(f1,5);(f2,5);(f300,5)} Flow Graph: [0->{},1->{2,3,4},2->{},3->{2,3},4->{2,4}] + 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. f2(A,B,C,D,E) -> f1(A,F,C,D,E) [1 >= A] (1,1) 1. f2(A,B,C,D,E) -> f300(A,B,C,D,E) [A >= 2 && C >= 2] (1,1) 2. f300(A,B,C,D,E) -> f1(A,F,C,D,E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0] (1,1) 3. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D] (?,1) 4. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] (?,1) Signature: {(f1,5);(f2,5);(f300,5)} Flow Graph: [0->{},1->{2,3,4},2->{},3->{2,3},4->{2,4}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = -1*x4 p(f2) = -1*x4 p(f300) = -1*x4 Following rules are strictly oriented: [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] ==> f300(A,B,C,D,E) = -1*D > -1 + -1*D = f300(A,B,C,1 + D,1 + E) Following rules are weakly oriented: [1 >= A] ==> f2(A,B,C,D,E) = -1*D >= -1*D = f1(A,F,C,D,E) [A >= 2 && C >= 2] ==> f2(A,B,C,D,E) = -1*D >= -1*D = f300(A,B,C,D,E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0] ==> f300(A,B,C,D,E) = -1*D >= -1*D = f1(A,F,C,D,E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D] ==> f300(A,B,C,D,E) = -1*D >= -1 + -1*D = f300(A,B,C,1 + D,1 + E) * Step 4: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f2(A,B,C,D,E) -> f1(A,F,C,D,E) [1 >= A] (1,1) 1. f2(A,B,C,D,E) -> f300(A,B,C,D,E) [A >= 2 && C >= 2] (1,1) 2. f300(A,B,C,D,E) -> f1(A,F,C,D,E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0] (1,1) 3. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D] (?,1) 4. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] (D,1) Signature: {(f1,5);(f2,5);(f300,5)} Flow Graph: [0->{},1->{2,3,4},2->{},3->{2,3},4->{2,4}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = -1*x4 p(f2) = -1*x4 p(f300) = -1*x4 Following rules are strictly oriented: [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D] ==> f300(A,B,C,D,E) = -1*D > -1 + -1*D = f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] ==> f300(A,B,C,D,E) = -1*D > -1 + -1*D = f300(A,B,C,1 + D,1 + E) Following rules are weakly oriented: [1 >= A] ==> f2(A,B,C,D,E) = -1*D >= -1*D = f1(A,F,C,D,E) [A >= 2 && C >= 2] ==> f2(A,B,C,D,E) = -1*D >= -1*D = f300(A,B,C,D,E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0] ==> f300(A,B,C,D,E) = -1*D >= -1*D = f1(A,F,C,D,E) * Step 5: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f2(A,B,C,D,E) -> f1(A,F,C,D,E) [1 >= A] (1,1) 1. f2(A,B,C,D,E) -> f300(A,B,C,D,E) [A >= 2 && C >= 2] (1,1) 2. f300(A,B,C,D,E) -> f1(A,F,C,D,E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0] (1,1) 3. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D] (D,1) 4. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] (D,1) Signature: {(f1,5);(f2,5);(f300,5)} Flow Graph: [0->{},1->{2,3,4},2->{},3->{2,3},4->{2,4}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))