YES * Step 1: TrivialSCCs YES + Considered Problem: Rules: 0. f12(A,B,C,D,E) -> f12(1 + A,B,C,D,E) [A >= 0 && 9 >= A] (?,1) 1. f25(A,B,C,D,E) -> f25(A,1 + B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && 9 >= B] (?,1) 2. f25(A,B,C,D,E) -> f36(A,B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && B >= 10] (?,1) 3. f12(A,B,C,D,E) -> f25(A,0,F,D,E) [A >= 0 && A >= 10] (?,1) 4. f0(A,B,C,D,E) -> f12(0,B,C,F,G) True (1,1) Signature: {(f0,5);(f12,5);(f25,5);(f36,5)} Flow Graph: [0->{0,3},1->{1,2},2->{},3->{1,2},4->{0,3}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths YES + Considered Problem: Rules: 0. f12(A,B,C,D,E) -> f12(1 + A,B,C,D,E) [A >= 0 && 9 >= A] (?,1) 1. f25(A,B,C,D,E) -> f25(A,1 + B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && 9 >= B] (?,1) 2. f25(A,B,C,D,E) -> f36(A,B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && B >= 10] (1,1) 3. f12(A,B,C,D,E) -> f25(A,0,F,D,E) [A >= 0 && A >= 10] (1,1) 4. f0(A,B,C,D,E) -> f12(0,B,C,F,G) True (1,1) Signature: {(f0,5);(f12,5);(f25,5);(f36,5)} Flow Graph: [0->{0,3},1->{1,2},2->{},3->{1,2},4->{0,3}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(3,2),(4,3)] * Step 3: Looptree YES + Considered Problem: Rules: 0. f12(A,B,C,D,E) -> f12(1 + A,B,C,D,E) [A >= 0 && 9 >= A] (?,1) 1. f25(A,B,C,D,E) -> f25(A,1 + B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && 9 >= B] (?,1) 2. f25(A,B,C,D,E) -> f36(A,B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && B >= 10] (1,1) 3. f12(A,B,C,D,E) -> f25(A,0,F,D,E) [A >= 0 && A >= 10] (1,1) 4. f0(A,B,C,D,E) -> f12(0,B,C,F,G) True (1,1) Signature: {(f0,5);(f12,5);(f25,5);(f36,5)} Flow Graph: [0->{0,3},1->{1,2},2->{},3->{1},4->{0}] + Applied Processor: Looptree + Details: We construct a looptree: P: [0,1,2,3,4] | +- p:[0] c: [0] | `- p:[1] c: [1] YES