YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + 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 WORST_CASE(?,O(n^1)) + 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: PolyRank WORST_CASE(?,O(n^1)) + 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: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = x3 p(f15) = -1*x2 + x3 p(f19) = -1*x2 + x3 p(f6) = -1*x2 + x3 Following rules are strictly oriented: [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B] ==> f6(A,B,C,D) = -1*B + C > -1 + -1*B + C = f6(2 + A,1 + B,C,D) Following rules are weakly oriented: True ==> f0(A,B,C,D) = C >= C = f6(0,0,C,D) [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B] ==> f6(A,B,C,D) = -1*B + C >= -1 + -1*B + C = f6(A,1 + B,C,D) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A = 1 + C] ==> f15(A,B,C,D) = -1*B + C >= -1*B + C = f19(1 + C,B,C,1) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && C >= A] ==> f15(A,B,C,D) = -1*B + C >= -1*B + C = f19(A,B,C,0) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A >= 2 + C] ==> f15(A,B,C,D) = -1*B + C >= -1*B + C = f19(A,B,C,0) [B >= 0 && A + B >= 0 && A >= 0 && B >= C && C >= 1 + A] ==> f6(A,B,C,D) = -1*B + C >= -1*B + C = f15(A,B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && A >= 1 + C && B >= C] ==> f6(A,B,C,D) = -1*B + C >= -1*B + C = f15(A,B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && B >= C && A = C] ==> f6(A,B,C,D) = -1*B + C >= A + -1*B = f19(A,B,A,1) * Step 4: PolyRank WORST_CASE(?,O(n^1)) + 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] (C,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: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = x3 p(f15) = -1*x2 + x3 p(f19) = -1*x2 + x3 p(f6) = -1*x2 + x3 Following rules are strictly oriented: [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B] ==> f6(A,B,C,D) = -1*B + C > -1 + -1*B + C = f6(A,1 + B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B] ==> f6(A,B,C,D) = -1*B + C > -1 + -1*B + C = f6(2 + A,1 + B,C,D) Following rules are weakly oriented: True ==> f0(A,B,C,D) = C >= C = f6(0,0,C,D) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A = 1 + C] ==> f15(A,B,C,D) = -1*B + C >= -1*B + C = f19(1 + C,B,C,1) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && C >= A] ==> f15(A,B,C,D) = -1*B + C >= -1*B + C = f19(A,B,C,0) [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A >= 2 + C] ==> f15(A,B,C,D) = -1*B + C >= -1*B + C = f19(A,B,C,0) [B >= 0 && A + B >= 0 && A >= 0 && B >= C && C >= 1 + A] ==> f6(A,B,C,D) = -1*B + C >= -1*B + C = f15(A,B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && A >= 1 + C && B >= C] ==> f6(A,B,C,D) = -1*B + C >= -1*B + C = f15(A,B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && B >= C && A = C] ==> f6(A,B,C,D) = -1*B + C >= A + -1*B = f19(A,B,A,1) * Step 5: KnowledgePropagation WORST_CASE(?,O(n^1)) + 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] (C,1) 2. f6(A,B,C,D) -> f6(2 + A,1 + B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B] (C,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: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))