YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f1(A,B,C) -> f2(A,B,C) True (1,1) 1. f2(A,B,C) -> f300(A,B,D) [B >= 3 && A >= 2] (?,1) 2. f2(A,B,C) -> f2(1 + A,1 + B,C) [2 >= B && A >= 2] (?,1) 3. f2(A,B,C) -> f2(1 + A,1 + B,C) [1 >= A] (?,1) Signature: {(f1,3);(f2,3);(f300,3)} Flow Graph: [0->{1,2,3},1->{},2->{1,2,3},3->{1,2,3}] + 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. f1(A,B,C) -> f2(A,B,C) True (1,1) 1. f2(A,B,C) -> f300(A,B,D) [B >= 3 && A >= 2] (1,1) 2. f2(A,B,C) -> f2(1 + A,1 + B,C) [2 >= B && A >= 2] (?,1) 3. f2(A,B,C) -> f2(1 + A,1 + B,C) [1 >= A] (?,1) Signature: {(f1,3);(f2,3);(f300,3)} Flow Graph: [0->{1,2,3},1->{},2->{1,2,3},3->{1,2,3}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(2,3)] * Step 3: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f1(A,B,C) -> f2(A,B,C) True (1,1) 1. f2(A,B,C) -> f300(A,B,D) [B >= 3 && A >= 2] (1,1) 2. f2(A,B,C) -> f2(1 + A,1 + B,C) [2 >= B && A >= 2] (?,1) 3. f2(A,B,C) -> f2(1 + A,1 + B,C) [1 >= A] (?,1) Signature: {(f1,3);(f2,3);(f300,3)} Flow Graph: [0->{1,2,3},1->{},2->{1,2},3->{1,2,3}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = 2 + -1*x1 p(f2) = 2 + -1*x1 p(f300) = 2 + -1*x1 Following rules are strictly oriented: [1 >= A] ==> f2(A,B,C) = 2 + -1*A > 1 + -1*A = f2(1 + A,1 + B,C) Following rules are weakly oriented: True ==> f1(A,B,C) = 2 + -1*A >= 2 + -1*A = f2(A,B,C) [B >= 3 && A >= 2] ==> f2(A,B,C) = 2 + -1*A >= 2 + -1*A = f300(A,B,D) [2 >= B && A >= 2] ==> f2(A,B,C) = 2 + -1*A >= 1 + -1*A = f2(1 + A,1 + B,C) * Step 4: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f1(A,B,C) -> f2(A,B,C) True (1,1) 1. f2(A,B,C) -> f300(A,B,D) [B >= 3 && A >= 2] (1,1) 2. f2(A,B,C) -> f2(1 + A,1 + B,C) [2 >= B && A >= 2] (?,1) 3. f2(A,B,C) -> f2(1 + A,1 + B,C) [1 >= A] (2 + A,1) Signature: {(f1,3);(f2,3);(f300,3)} Flow Graph: [0->{1,2,3},1->{},2->{1,2},3->{1,2,3}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f1) = 3 + -1*x2 p(f2) = 3 + -1*x2 p(f300) = 3 + -1*x2 Following rules are strictly oriented: [2 >= B && A >= 2] ==> f2(A,B,C) = 3 + -1*B > 2 + -1*B = f2(1 + A,1 + B,C) Following rules are weakly oriented: True ==> f1(A,B,C) = 3 + -1*B >= 3 + -1*B = f2(A,B,C) [B >= 3 && A >= 2] ==> f2(A,B,C) = 3 + -1*B >= 3 + -1*B = f300(A,B,D) [1 >= A] ==> f2(A,B,C) = 3 + -1*B >= 2 + -1*B = f2(1 + A,1 + B,C) * Step 5: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f1(A,B,C) -> f2(A,B,C) True (1,1) 1. f2(A,B,C) -> f300(A,B,D) [B >= 3 && A >= 2] (1,1) 2. f2(A,B,C) -> f2(1 + A,1 + B,C) [2 >= B && A >= 2] (3 + B,1) 3. f2(A,B,C) -> f2(1 + A,1 + B,C) [1 >= A] (2 + A,1) Signature: {(f1,3);(f2,3);(f300,3)} Flow Graph: [0->{1,2,3},1->{},2->{1,2},3->{1,2,3}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))