MAYBE * Step 1: TrivialSCCs MAYBE + Considered Problem: Rules: 0. f0(A,B) -> f3(0,B) True (1,1) 1. f3(A,B) -> f3(1 + C,B) [A = 5] (?,1) 2. f3(A,B) -> f3(1 + A,A) [9 >= A && 4 >= A] (?,1) 3. f3(A,B) -> f3(1 + A,A) [9 >= A && A >= 6] (?,1) 4. f3(A,B) -> f12(A,B) [A >= 10] (?,1) Signature: {(f0,2);(f12,2);(f3,2)} Flow Graph: [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3,4},4->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths MAYBE + Considered Problem: Rules: 0. f0(A,B) -> f3(0,B) True (1,1) 1. f3(A,B) -> f3(1 + C,B) [A = 5] (?,1) 2. f3(A,B) -> f3(1 + A,A) [9 >= A && 4 >= A] (?,1) 3. f3(A,B) -> f3(1 + A,A) [9 >= A && A >= 6] (?,1) 4. f3(A,B) -> f12(A,B) [A >= 10] (1,1) Signature: {(f0,2);(f12,2);(f3,2)} Flow Graph: [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3,4},4->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(0,1),(0,3),(0,4),(2,3),(2,4),(3,1),(3,2)] * Step 3: AddSinks MAYBE + Considered Problem: Rules: 0. f0(A,B) -> f3(0,B) True (1,1) 1. f3(A,B) -> f3(1 + C,B) [A = 5] (?,1) 2. f3(A,B) -> f3(1 + A,A) [9 >= A && 4 >= A] (?,1) 3. f3(A,B) -> f3(1 + A,A) [9 >= A && A >= 6] (?,1) 4. f3(A,B) -> f12(A,B) [A >= 10] (1,1) Signature: {(f0,2);(f12,2);(f3,2)} Flow Graph: [0->{2},1->{1,2,3,4},2->{1,2},3->{3,4},4->{}] + Applied Processor: AddSinks + Details: () * Step 4: UnsatPaths MAYBE + Considered Problem: Rules: 0. f0(A,B) -> f3(0,B) True (1,1) 1. f3(A,B) -> f3(1 + C,B) [A = 5] (?,1) 2. f3(A,B) -> f3(1 + A,A) [9 >= A && 4 >= A] (?,1) 3. f3(A,B) -> f3(1 + A,A) [9 >= A && A >= 6] (?,1) 4. f3(A,B) -> f12(A,B) [A >= 10] (?,1) 5. f3(A,B) -> exitus616(A,B) True (?,1) Signature: {(exitus616,2);(f0,2);(f12,2);(f3,2)} Flow Graph: [0->{1,2,3,4,5},1->{1,2,3,4,5},2->{1,2,3,4,5},3->{1,2,3,4,5},4->{},5->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(0,1),(0,3),(0,4),(2,3),(2,4),(3,1),(3,2)] * Step 5: Failure MAYBE + Considered Problem: Rules: 0. f0(A,B) -> f3(0,B) True (1,1) 1. f3(A,B) -> f3(1 + C,B) [A = 5] (?,1) 2. f3(A,B) -> f3(1 + A,A) [9 >= A && 4 >= A] (?,1) 3. f3(A,B) -> f3(1 + A,A) [9 >= A && A >= 6] (?,1) 4. f3(A,B) -> f12(A,B) [A >= 10] (?,1) 5. f3(A,B) -> exitus616(A,B) True (?,1) Signature: {(exitus616,2);(f0,2);(f12,2);(f3,2)} Flow Graph: [0->{2,5},1->{1,2,3,4,5},2->{1,2,5},3->{3,4,5},4->{},5->{}] + Applied Processor: LooptreeTransformer + Details: We construct a looptree: P: [0,1,2,3,4,5] | +- p:[2,1] c: [] | `- p:[3] c: [3] MAYBE