YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f4(A,B,C) -> f5(A,B,1) [0 >= A && 0 >= B] (?,1) 1. f0(A,B,C) -> f2(A,B,1) [A >= 1] (1,1) 2. f4(A,B,C) -> f4(A,-1 + B,C) [B >= 1] (?,1) 3. f0(A,B,C) -> f4(A,B,0) [0 >= A] (1,1) Signature: {(f0,3);(f2,3);(f4,3);(f5,3)} Flow Graph: [0->{},1->{},2->{0,2},3->{0,2}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f4(A,B,C) -> f5(A,B,1) [0 >= A && 0 >= B] (1,1) 1. f0(A,B,C) -> f2(A,B,1) [A >= 1] (1,1) 2. f4(A,B,C) -> f4(A,-1 + B,C) [B >= 1] (?,1) 3. f0(A,B,C) -> f4(A,B,0) [0 >= A] (1,1) Signature: {(f0,3);(f2,3);(f4,3);(f5,3)} Flow Graph: [0->{},1->{},2->{0,2},3->{0,2}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = x2 p(f2) = x2 p(f4) = x2 p(f5) = x2 Following rules are strictly oriented: [B >= 1] ==> f4(A,B,C) = B > -1 + B = f4(A,-1 + B,C) Following rules are weakly oriented: [0 >= A && 0 >= B] ==> f4(A,B,C) = B >= B = f5(A,B,1) [A >= 1] ==> f0(A,B,C) = B >= B = f2(A,B,1) [0 >= A] ==> f0(A,B,C) = B >= B = f4(A,B,0) * Step 3: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. f4(A,B,C) -> f5(A,B,1) [0 >= A && 0 >= B] (1,1) 1. f0(A,B,C) -> f2(A,B,1) [A >= 1] (1,1) 2. f4(A,B,C) -> f4(A,-1 + B,C) [B >= 1] (B,1) 3. f0(A,B,C) -> f4(A,B,0) [0 >= A] (1,1) Signature: {(f0,3);(f2,3);(f4,3);(f5,3)} Flow Graph: [0->{},1->{},2->{0,2},3->{0,2}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))