NO * Step 1: TrivialSCCs NO + Considered Problem: Rules: 0. f4(A,B,C,D,E,F) -> f0(G,0,0,H,G,G) True (1,1) 1. f4(A,B,C,D,E,F) -> f2(1 + H,G,G,D,H,H) [0 >= 1 + G && H >= 1] (1,1) 2. f4(A,B,C,D,E,F) -> f2(1 + H,G,G,D,H,H) [G >= 1 && H >= 1] (1,1) 3. f2(A,B,C,D,E,F) -> f0(A,0,0,G,E,F) [-1 + A + -1*F >= 0 && -1 + F >= 0 && -3 + A + F >= 0 && -2 + A >= 0] (?,1) 4. f2(A,B,C,D,E,F) -> f2(1 + A,G,G,D,E,F) [-1 + A + -1*F >= 0 && -1 + F >= 0 && -3 + A + F >= 0 && -2 + A >= 0 && 0 >= 1 + G && A >= 1] (?,1) 5. f2(A,B,C,D,E,F) -> f2(1 + A,G,G,D,E,F) [-1 + A + -1*F >= 0 && -1 + F >= 0 && -3 + A + F >= 0 && -2 + A >= 0 && G >= 1 && A >= 1] (?,1) Signature: {(f0,6);(f2,6);(f4,6)} Flow Graph: [0->{},1->{3,4,5},2->{3,4,5},3->{},4->{3,4,5},5->{3,4,5}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: Looptree NO + Considered Problem: Rules: 0. f4(A,B,C,D,E,F) -> f0(G,0,0,H,G,G) True (1,1) 1. f4(A,B,C,D,E,F) -> f2(1 + H,G,G,D,H,H) [0 >= 1 + G && H >= 1] (1,1) 2. f4(A,B,C,D,E,F) -> f2(1 + H,G,G,D,H,H) [G >= 1 && H >= 1] (1,1) 3. f2(A,B,C,D,E,F) -> f0(A,0,0,G,E,F) [-1 + A + -1*F >= 0 && -1 + F >= 0 && -3 + A + F >= 0 && -2 + A >= 0] (1,1) 4. f2(A,B,C,D,E,F) -> f2(1 + A,G,G,D,E,F) [-1 + A + -1*F >= 0 && -1 + F >= 0 && -3 + A + F >= 0 && -2 + A >= 0 && 0 >= 1 + G && A >= 1] (?,1) 5. f2(A,B,C,D,E,F) -> f2(1 + A,G,G,D,E,F) [-1 + A + -1*F >= 0 && -1 + F >= 0 && -3 + A + F >= 0 && -2 + A >= 0 && G >= 1 && A >= 1] (?,1) Signature: {(f0,6);(f2,6);(f4,6)} Flow Graph: [0->{},1->{3,4,5},2->{3,4,5},3->{},4->{3,4,5},5->{3,4,5}] + Applied Processor: Looptree + Details: We construct a looptree: P: [0,1,2,3,4,5] | `- p:[4,5] c: [] NO