YES * Step 1: TrivialSCCs YES + Considered Problem: Rules: 0. evalndloopstart(A) -> evalndloopentryin(A) True (1,1) 1. evalndloopentryin(A) -> evalndloopbbin(0) True (?,1) 2. evalndloopbbin(A) -> evalndloopbbin(B) [A >= 0 && 2 + A >= B && B >= 1 + A && 9 >= B] (?,1) 3. evalndloopbbin(A) -> evalndloopreturnin(A) [A >= 0 && B >= 3 + A] (?,1) 4. evalndloopbbin(A) -> evalndloopreturnin(A) [A >= 0 && A >= B] (?,1) 5. evalndloopbbin(A) -> evalndloopreturnin(A) [A >= 0 && B >= 10] (?,1) 6. evalndloopreturnin(A) -> evalndloopstop(A) [A >= 0] (?,1) Signature: {(evalndloopbbin,1);(evalndloopentryin,1);(evalndloopreturnin,1);(evalndloopstart,1);(evalndloopstop,1)} Flow Graph: [0->{1},1->{2,3,4,5},2->{2,3,4,5},3->{6},4->{6},5->{6},6->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: Looptree YES + Considered Problem: Rules: 0. evalndloopstart(A) -> evalndloopentryin(A) True (1,1) 1. evalndloopentryin(A) -> evalndloopbbin(0) True (1,1) 2. evalndloopbbin(A) -> evalndloopbbin(B) [A >= 0 && 2 + A >= B && B >= 1 + A && 9 >= B] (?,1) 3. evalndloopbbin(A) -> evalndloopreturnin(A) [A >= 0 && B >= 3 + A] (1,1) 4. evalndloopbbin(A) -> evalndloopreturnin(A) [A >= 0 && A >= B] (1,1) 5. evalndloopbbin(A) -> evalndloopreturnin(A) [A >= 0 && B >= 10] (1,1) 6. evalndloopreturnin(A) -> evalndloopstop(A) [A >= 0] (1,1) Signature: {(evalndloopbbin,1);(evalndloopentryin,1);(evalndloopreturnin,1);(evalndloopstart,1);(evalndloopstop,1)} Flow Graph: [0->{1},1->{2,3,4,5},2->{2,3,4,5},3->{6},4->{6},5->{6},6->{}] + Applied Processor: Looptree + Details: We construct a looptree: P: [0,1,2,3,4,5,6] | `- p:[2] c: [2] YES