YES * Step 1: TrivialSCCs YES + Considered Problem: Rules: 0. f0(A,B,C,D) -> f6(0,0,C,D) True (1,1) 1. f6(A,B,C,D) -> f6(A,1 + B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B] (?,1) 2. f6(A,B,C,D) -> f6(2 + A,1 + B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B] (?,1) 3. f15(A,B,C,D) -> f19(1 + C,B,C,1) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A = 1 + C] (?,1) 4. f15(A,B,C,D) -> f19(A,B,C,0) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && C >= A] (?,1) 5. f15(A,B,C,D) -> f19(A,B,C,0) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A >= 2 + C] (?,1) 6. f6(A,B,C,D) -> f15(A,B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && B >= C && C >= 1 + A] (?,1) 7. f6(A,B,C,D) -> f15(A,B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && A >= 1 + C && B >= C] (?,1) 8. f6(A,B,C,D) -> f19(A,B,A,1) [B >= 0 && A + B >= 0 && A >= 0 && B >= C && A = C] (?,1) Signature: {(f0,4);(f15,4);(f19,4);(f6,4)} Flow Graph: [0->{1,2,6,7,8},1->{1,2,6,7,8},2->{1,2,6,7,8},3->{},4->{},5->{},6->{3,4,5},7->{3,4,5},8->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths YES + Considered Problem: Rules: 0. f0(A,B,C,D) -> f6(0,0,C,D) True (1,1) 1. f6(A,B,C,D) -> f6(A,1 + B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B] (?,1) 2. f6(A,B,C,D) -> f6(2 + A,1 + B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B] (?,1) 3. f15(A,B,C,D) -> f19(1 + C,B,C,1) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A = 1 + C] (1,1) 4. f15(A,B,C,D) -> f19(A,B,C,0) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && C >= A] (1,1) 5. f15(A,B,C,D) -> f19(A,B,C,0) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A >= 2 + C] (1,1) 6. f6(A,B,C,D) -> f15(A,B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && B >= C && C >= 1 + A] (1,1) 7. f6(A,B,C,D) -> f15(A,B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && A >= 1 + C && B >= C] (1,1) 8. f6(A,B,C,D) -> f19(A,B,A,1) [B >= 0 && A + B >= 0 && A >= 0 && B >= C && A = C] (1,1) Signature: {(f0,4);(f15,4);(f19,4);(f6,4)} Flow Graph: [0->{1,2,6,7,8},1->{1,2,6,7,8},2->{1,2,6,7,8},3->{},4->{},5->{},6->{3,4,5},7->{3,4,5},8->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(0,6),(6,3),(6,5),(7,4)] * Step 3: Looptree YES + Considered Problem: Rules: 0. f0(A,B,C,D) -> f6(0,0,C,D) True (1,1) 1. f6(A,B,C,D) -> f6(A,1 + B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B] (?,1) 2. f6(A,B,C,D) -> f6(2 + A,1 + B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B] (?,1) 3. f15(A,B,C,D) -> f19(1 + C,B,C,1) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A = 1 + C] (1,1) 4. f15(A,B,C,D) -> f19(A,B,C,0) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && C >= A] (1,1) 5. f15(A,B,C,D) -> f19(A,B,C,0) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A >= 2 + C] (1,1) 6. f6(A,B,C,D) -> f15(A,B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && B >= C && C >= 1 + A] (1,1) 7. f6(A,B,C,D) -> f15(A,B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && A >= 1 + C && B >= C] (1,1) 8. f6(A,B,C,D) -> f19(A,B,A,1) [B >= 0 && A + B >= 0 && A >= 0 && B >= C && A = C] (1,1) Signature: {(f0,4);(f15,4);(f19,4);(f6,4)} Flow Graph: [0->{1,2,7,8},1->{1,2,6,7,8},2->{1,2,6,7,8},3->{},4->{},5->{},6->{4},7->{3,5},8->{}] + Applied Processor: Looptree + Details: We construct a looptree: P: [0,1,2,3,4,5,6,7,8] | `- p:[1,2] c: [2] | `- p:[1] c: [1] YES