MAYBE * Step 1: TrivialSCCs MAYBE + Considered Problem: Rules: 0. f2(A,B,C) -> f300(A,B,D) [0 >= 1 + A] (?,1) 1. f2(A,B,C) -> f2(A,0,C) [A >= 0] (?,1) 2. f2(A,B,C) -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 0] (?,1) 3. f2(A,B,C) -> f2(-1 + A,D,C) [D >= 1 && A >= 0] (?,1) 4. f3(A,B,C) -> f2(A,B,C) True (1,1) Signature: {(f2,3);(f3,3);(f300,3)} Flow Graph: [0->{},1->{0,1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths MAYBE + Considered Problem: Rules: 0. f2(A,B,C) -> f300(A,B,D) [0 >= 1 + A] (1,1) 1. f2(A,B,C) -> f2(A,0,C) [A >= 0] (?,1) 2. f2(A,B,C) -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 0] (?,1) 3. f2(A,B,C) -> f2(-1 + A,D,C) [D >= 1 && A >= 0] (?,1) 4. f3(A,B,C) -> f2(A,B,C) True (1,1) Signature: {(f2,3);(f3,3);(f300,3)} Flow Graph: [0->{},1->{0,1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(1,0)] * Step 3: AddSinks MAYBE + Considered Problem: Rules: 0. f2(A,B,C) -> f300(A,B,D) [0 >= 1 + A] (1,1) 1. f2(A,B,C) -> f2(A,0,C) [A >= 0] (?,1) 2. f2(A,B,C) -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 0] (?,1) 3. f2(A,B,C) -> f2(-1 + A,D,C) [D >= 1 && A >= 0] (?,1) 4. f3(A,B,C) -> f2(A,B,C) True (1,1) Signature: {(f2,3);(f3,3);(f300,3)} Flow Graph: [0->{},1->{1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}] + Applied Processor: AddSinks + Details: () * Step 4: UnsatPaths MAYBE + Considered Problem: Rules: 0. f2(A,B,C) -> f300(A,B,D) [0 >= 1 + A] (?,1) 1. f2(A,B,C) -> f2(A,0,C) [A >= 0] (?,1) 2. f2(A,B,C) -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 0] (?,1) 3. f2(A,B,C) -> f2(-1 + A,D,C) [D >= 1 && A >= 0] (?,1) 4. f3(A,B,C) -> f2(A,B,C) True (1,1) 5. f2(A,B,C) -> exitus616(A,B,C) True (?,1) Signature: {(exitus616,3);(f2,3);(f3,3);(f300,3)} Flow Graph: [0->{},1->{0,1,2,3,5},2->{0,1,2,3,5},3->{0,1,2,3,5},4->{0,1,2,3,5},5->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(1,0)] * Step 5: Failure MAYBE + Considered Problem: Rules: 0. f2(A,B,C) -> f300(A,B,D) [0 >= 1 + A] (?,1) 1. f2(A,B,C) -> f2(A,0,C) [A >= 0] (?,1) 2. f2(A,B,C) -> f2(-1 + A,D,C) [0 >= 1 + D && A >= 0] (?,1) 3. f2(A,B,C) -> f2(-1 + A,D,C) [D >= 1 && A >= 0] (?,1) 4. f3(A,B,C) -> f2(A,B,C) True (1,1) 5. f2(A,B,C) -> exitus616(A,B,C) True (?,1) Signature: {(exitus616,3);(f2,3);(f3,3);(f300,3)} Flow Graph: [0->{},1->{1,2,3,5},2->{0,1,2,3,5},3->{0,1,2,3,5},4->{0,1,2,3,5},5->{}] + Applied Processor: LooptreeTransformer + Details: We construct a looptree: P: [0,1,2,3,4,5] | `- p:[1,2,3] c: [3] | `- p:[1,2] c: [2] | `- p:[1] c: [] MAYBE