YES * Step 1: TrivialSCCs YES + Considered Problem: Rules: 0. l0(A,B,C,D) -> l1(0,B,C,D) True (1,1) 1. l1(A,B,C,D) -> l1(1 + A,-1 + B,C,D) [A >= 0 && B >= 1] (?,1) 2. l1(A,B,C,D) -> l2(A,B,A,D) [A >= 0 && 0 >= B] (?,1) 3. l2(A,B,C,D) -> l3(A,B,C,C) [A + -1*C >= 0 && C >= 0 && -1*B + C >= 0 && A + C >= 0 && -1*B >= 0 && A + -1*B >= 0 && A >= 0 && C >= 1] (?,1) 4. l3(A,B,C,D) -> l3(A,B,C,-1 + D) [C + -1*D >= 0 (?,1) && A + -1*D >= 0 && A + -1*C >= 0 && -1 + C >= 0 && -1 + -1*B + C >= 0 && -2 + A + C >= 0 && -1*B >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && D >= 1 && C >= 1] 5. l3(A,B,C,D) -> l2(A,B,-1 + C,D) [C + -1*D >= 0 (?,1) && A + -1*D >= 0 && A + -1*C >= 0 && -1 + C >= 0 && -1 + -1*B + C >= 0 && -2 + A + C >= 0 && -1*B >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && 0 >= D && C >= 1] Signature: {(l0,4);(l1,4);(l2,4);(l3,4)} Flow Graph: [0->{1,2},1->{1,2},2->{3},3->{4,5},4->{4,5},5->{3}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths YES + Considered Problem: Rules: 0. l0(A,B,C,D) -> l1(0,B,C,D) True (1,1) 1. l1(A,B,C,D) -> l1(1 + A,-1 + B,C,D) [A >= 0 && B >= 1] (?,1) 2. l1(A,B,C,D) -> l2(A,B,A,D) [A >= 0 && 0 >= B] (1,1) 3. l2(A,B,C,D) -> l3(A,B,C,C) [A + -1*C >= 0 && C >= 0 && -1*B + C >= 0 && A + C >= 0 && -1*B >= 0 && A + -1*B >= 0 && A >= 0 && C >= 1] (?,1) 4. l3(A,B,C,D) -> l3(A,B,C,-1 + D) [C + -1*D >= 0 (?,1) && A + -1*D >= 0 && A + -1*C >= 0 && -1 + C >= 0 && -1 + -1*B + C >= 0 && -2 + A + C >= 0 && -1*B >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && D >= 1 && C >= 1] 5. l3(A,B,C,D) -> l2(A,B,-1 + C,D) [C + -1*D >= 0 (?,1) && A + -1*D >= 0 && A + -1*C >= 0 && -1 + C >= 0 && -1 + -1*B + C >= 0 && -2 + A + C >= 0 && -1*B >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && 0 >= D && C >= 1] Signature: {(l0,4);(l1,4);(l2,4);(l3,4)} Flow Graph: [0->{1,2},1->{1,2},2->{3},3->{4,5},4->{4,5},5->{3}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(3,5)] * Step 3: Looptree YES + Considered Problem: Rules: 0. l0(A,B,C,D) -> l1(0,B,C,D) True (1,1) 1. l1(A,B,C,D) -> l1(1 + A,-1 + B,C,D) [A >= 0 && B >= 1] (?,1) 2. l1(A,B,C,D) -> l2(A,B,A,D) [A >= 0 && 0 >= B] (1,1) 3. l2(A,B,C,D) -> l3(A,B,C,C) [A + -1*C >= 0 && C >= 0 && -1*B + C >= 0 && A + C >= 0 && -1*B >= 0 && A + -1*B >= 0 && A >= 0 && C >= 1] (?,1) 4. l3(A,B,C,D) -> l3(A,B,C,-1 + D) [C + -1*D >= 0 (?,1) && A + -1*D >= 0 && A + -1*C >= 0 && -1 + C >= 0 && -1 + -1*B + C >= 0 && -2 + A + C >= 0 && -1*B >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && D >= 1 && C >= 1] 5. l3(A,B,C,D) -> l2(A,B,-1 + C,D) [C + -1*D >= 0 (?,1) && A + -1*D >= 0 && A + -1*C >= 0 && -1 + C >= 0 && -1 + -1*B + C >= 0 && -2 + A + C >= 0 && -1*B >= 0 && -1 + A + -1*B >= 0 && -1 + A >= 0 && 0 >= D && C >= 1] Signature: {(l0,4);(l1,4);(l2,4);(l3,4)} Flow Graph: [0->{1,2},1->{1,2},2->{3},3->{4},4->{4,5},5->{3}] + Applied Processor: Looptree + Details: We construct a looptree: P: [0,1,2,3,4,5] | +- p:[1] c: [1] | `- p:[3,5,4] c: [5] | `- p:[4] c: [4] YES