YES(?,O(1)) * Step 1: UnsatPaths WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f12(A,B,C,D,E) -> f12(1 + A,B,C,D,E) [A >= 0 && 9 >= A] (?,1) 1. f25(A,B,C,D,E) -> f25(A,1 + B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && 9 >= B] (?,1) 2. f25(A,B,C,D,E) -> f36(A,B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && B >= 10] (?,1) 3. f12(A,B,C,D,E) -> f25(A,0,F,D,E) [A >= 0 && A >= 10] (?,1) 4. f0(A,B,C,D,E) -> f12(0,B,C,F,G) True (1,1) Signature: {(f0,5);(f12,5);(f25,5);(f36,5)} Flow Graph: [0->{0,3},1->{1,2},2->{},3->{1,2},4->{0,3}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(3,2),(4,3)] * Step 2: TrivialSCCs WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f12(A,B,C,D,E) -> f12(1 + A,B,C,D,E) [A >= 0 && 9 >= A] (?,1) 1. f25(A,B,C,D,E) -> f25(A,1 + B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && 9 >= B] (?,1) 2. f25(A,B,C,D,E) -> f36(A,B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && B >= 10] (?,1) 3. f12(A,B,C,D,E) -> f25(A,0,F,D,E) [A >= 0 && A >= 10] (?,1) 4. f0(A,B,C,D,E) -> f12(0,B,C,F,G) True (1,1) Signature: {(f0,5);(f12,5);(f25,5);(f36,5)} Flow Graph: [0->{0,3},1->{1,2},2->{},3->{1},4->{0}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 3: PolyRank WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f12(A,B,C,D,E) -> f12(1 + A,B,C,D,E) [A >= 0 && 9 >= A] (?,1) 1. f25(A,B,C,D,E) -> f25(A,1 + B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && 9 >= B] (?,1) 2. f25(A,B,C,D,E) -> f36(A,B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && B >= 10] (1,1) 3. f12(A,B,C,D,E) -> f25(A,0,F,D,E) [A >= 0 && A >= 10] (1,1) 4. f0(A,B,C,D,E) -> f12(0,B,C,F,G) True (1,1) Signature: {(f0,5);(f12,5);(f25,5);(f36,5)} Flow Graph: [0->{0,3},1->{1,2},2->{},3->{1},4->{0}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = 10 p(f12) = 10 p(f25) = 10 + -1*x2 p(f36) = 10 + -1*x2 Following rules are strictly oriented: [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && 9 >= B] ==> f25(A,B,C,D,E) = 10 + -1*B > 9 + -1*B = f25(A,1 + B,C,D,E) Following rules are weakly oriented: [A >= 0 && 9 >= A] ==> f12(A,B,C,D,E) = 10 >= 10 = f12(1 + A,B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && B >= 10] ==> f25(A,B,C,D,E) = 10 + -1*B >= 10 + -1*B = f36(A,B,C,D,E) [A >= 0 && A >= 10] ==> f12(A,B,C,D,E) = 10 >= 10 = f25(A,0,F,D,E) True ==> f0(A,B,C,D,E) = 10 >= 10 = f12(0,B,C,F,G) * Step 4: PolyRank WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f12(A,B,C,D,E) -> f12(1 + A,B,C,D,E) [A >= 0 && 9 >= A] (?,1) 1. f25(A,B,C,D,E) -> f25(A,1 + B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && 9 >= B] (10,1) 2. f25(A,B,C,D,E) -> f36(A,B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && B >= 10] (1,1) 3. f12(A,B,C,D,E) -> f25(A,0,F,D,E) [A >= 0 && A >= 10] (1,1) 4. f0(A,B,C,D,E) -> f12(0,B,C,F,G) True (1,1) Signature: {(f0,5);(f12,5);(f25,5);(f36,5)} Flow Graph: [0->{0,3},1->{1,2},2->{},3->{1},4->{0}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = 10 p(f12) = 10 + -1*x1 p(f25) = 10 + -1*x1 p(f36) = 10 + -1*x1 Following rules are strictly oriented: [A >= 0 && 9 >= A] ==> f12(A,B,C,D,E) = 10 + -1*A > 9 + -1*A = f12(1 + A,B,C,D,E) Following rules are weakly oriented: [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && 9 >= B] ==> f25(A,B,C,D,E) = 10 + -1*A >= 10 + -1*A = f25(A,1 + B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && B >= 10] ==> f25(A,B,C,D,E) = 10 + -1*A >= 10 + -1*A = f36(A,B,C,D,E) [A >= 0 && A >= 10] ==> f12(A,B,C,D,E) = 10 + -1*A >= 10 + -1*A = f25(A,0,F,D,E) True ==> f0(A,B,C,D,E) = 10 >= 10 = f12(0,B,C,F,G) * Step 5: KnowledgePropagation WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f12(A,B,C,D,E) -> f12(1 + A,B,C,D,E) [A >= 0 && 9 >= A] (10,1) 1. f25(A,B,C,D,E) -> f25(A,1 + B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && 9 >= B] (10,1) 2. f25(A,B,C,D,E) -> f36(A,B,C,D,E) [B >= 0 && -10 + A + B >= 0 && -10 + A >= 0 && B >= 10] (1,1) 3. f12(A,B,C,D,E) -> f25(A,0,F,D,E) [A >= 0 && A >= 10] (1,1) 4. f0(A,B,C,D,E) -> f12(0,B,C,F,G) True (1,1) Signature: {(f0,5);(f12,5);(f25,5);(f36,5)} Flow Graph: [0->{0,3},1->{1,2},2->{},3->{1},4->{0}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(1))