MAYBE * Step 1: TrivialSCCs MAYBE + Considered Problem: Rules: 0. f0(A,B,C,D,E) -> f3(0,0,C,D,E) True (1,1) 1. f3(A,B,C,D,E) -> f3(A,B,-1 + C,F,E) [C >= 1 && F >= 1] (?,1) 2. f3(A,B,C,D,E) -> f3(A,B,-2 + C,F,E) [C >= 1 && 0 >= F] (?,1) 3. f3(A,B,C,D,E) -> f6(A,B,C,D,F) [0 >= C] (?,1) 4. f6(A,B,C,D,E) -> f6(1,B,C,D,F) [E >= 1] (?,1) 5. f6(A,B,C,D,E) -> f6(0,B,C,D,F) [0 >= E] (?,1) Signature: {(f0,5);(f3,5);(f6,5)} Flow Graph: [0->{1,2,3},1->{1,2,3},2->{1,2,3},3->{4,5},4->{4,5},5->{4,5}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: AddSinks MAYBE + Considered Problem: Rules: 0. f0(A,B,C,D,E) -> f3(0,0,C,D,E) True (1,1) 1. f3(A,B,C,D,E) -> f3(A,B,-1 + C,F,E) [C >= 1 && F >= 1] (?,1) 2. f3(A,B,C,D,E) -> f3(A,B,-2 + C,F,E) [C >= 1 && 0 >= F] (?,1) 3. f3(A,B,C,D,E) -> f6(A,B,C,D,F) [0 >= C] (1,1) 4. f6(A,B,C,D,E) -> f6(1,B,C,D,F) [E >= 1] (?,1) 5. f6(A,B,C,D,E) -> f6(0,B,C,D,F) [0 >= E] (?,1) Signature: {(f0,5);(f3,5);(f6,5)} Flow Graph: [0->{1,2,3},1->{1,2,3},2->{1,2,3},3->{4,5},4->{4,5},5->{4,5}] + Applied Processor: AddSinks + Details: () * Step 3: Failure MAYBE + Considered Problem: Rules: 0. f0(A,B,C,D,E) -> f3(0,0,C,D,E) True (1,1) 1. f3(A,B,C,D,E) -> f3(A,B,-1 + C,F,E) [C >= 1 && F >= 1] (?,1) 2. f3(A,B,C,D,E) -> f3(A,B,-2 + C,F,E) [C >= 1 && 0 >= F] (?,1) 3. f3(A,B,C,D,E) -> f6(A,B,C,D,F) [0 >= C] (?,1) 4. f6(A,B,C,D,E) -> f6(1,B,C,D,F) [E >= 1] (?,1) 5. f6(A,B,C,D,E) -> f6(0,B,C,D,F) [0 >= E] (?,1) 6. f6(A,B,C,D,E) -> exitus616(A,B,C,D,E) True (?,1) Signature: {(exitus616,5);(f0,5);(f3,5);(f6,5)} Flow Graph: [0->{1,2,3},1->{1,2,3},2->{1,2,3},3->{4,5,6},4->{4,5,6},5->{4,5,6},6->{}] + Applied Processor: LooptreeTransformer + Details: We construct a looptree: P: [0,1,2,3,4,5,6] | +- p:[1,2] c: [2] | | | `- p:[1] c: [1] | `- p:[4,5] c: [] MAYBE