YES(?,O(1)) * Step 1: TrivialSCCs WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B,C,D) -> f19(A,999,C,1) True (1,1) 1. f19(A,B,C,D) -> f19(A,-1 + B,C,D) [1 + -1*D >= 0 (?,1) && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && B >= 0] 2. f19(A,B,C,D) -> f28(A,B,999,D) [1 + -1*D >= 0 (?,1) && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && 0 >= 1 + B] 3. f28(A,B,C,D) -> f28(A,B,-1 + C,D) [1 + -1*D >= 0 (?,1) && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && C >= 0] 4. f28(A,B,C,D) -> f36(A,B,C,D) [1 + -1*D >= 0 (?,1) && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && 0 >= 1 + C] Signature: {(f0,4);(f19,4);(f28,4);(f36,4)} Flow Graph: [0->{1,2},1->{1,2},2->{3,4},3->{3,4},4->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B,C,D) -> f19(A,999,C,1) True (1,1) 1. f19(A,B,C,D) -> f19(A,-1 + B,C,D) [1 + -1*D >= 0 (?,1) && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && B >= 0] 2. f19(A,B,C,D) -> f28(A,B,999,D) [1 + -1*D >= 0 (1,1) && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && 0 >= 1 + B] 3. f28(A,B,C,D) -> f28(A,B,-1 + C,D) [1 + -1*D >= 0 (?,1) && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && C >= 0] 4. f28(A,B,C,D) -> f36(A,B,C,D) [1 + -1*D >= 0 (1,1) && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && 0 >= 1 + C] Signature: {(f0,4);(f19,4);(f28,4);(f36,4)} Flow Graph: [0->{1,2},1->{1,2},2->{3,4},3->{3,4},4->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(0,2),(2,4)] * Step 3: PolyRank WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B,C,D) -> f19(A,999,C,1) True (1,1) 1. f19(A,B,C,D) -> f19(A,-1 + B,C,D) [1 + -1*D >= 0 (?,1) && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && B >= 0] 2. f19(A,B,C,D) -> f28(A,B,999,D) [1 + -1*D >= 0 (1,1) && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && 0 >= 1 + B] 3. f28(A,B,C,D) -> f28(A,B,-1 + C,D) [1 + -1*D >= 0 (?,1) && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && C >= 0] 4. f28(A,B,C,D) -> f36(A,B,C,D) [1 + -1*D >= 0 (1,1) && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && 0 >= 1 + C] Signature: {(f0,4);(f19,4);(f28,4);(f36,4)} Flow Graph: [0->{1},1->{1,2},2->{3},3->{3,4},4->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = 1000 p(f19) = 1000*x4 p(f28) = -999 + x3 + 1000*x4 p(f36) = -999 + x3 + 1000*x4 Following rules are strictly oriented: [1 + -1*D >= 0 ==> && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && C >= 0] f28(A,B,C,D) = -999 + C + 1000*D > -1000 + C + 1000*D = f28(A,B,-1 + C,D) Following rules are weakly oriented: True ==> f0(A,B,C,D) = 1000 >= 1000 = f19(A,999,C,1) [1 + -1*D >= 0 ==> && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && B >= 0] f19(A,B,C,D) = 1000*D >= 1000*D = f19(A,-1 + B,C,D) [1 + -1*D >= 0 ==> && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && 0 >= 1 + B] f19(A,B,C,D) = 1000*D >= 1000*D = f28(A,B,999,D) [1 + -1*D >= 0 ==> && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && 0 >= 1 + C] f28(A,B,C,D) = -999 + C + 1000*D >= -999 + C + 1000*D = f36(A,B,C,D) * Step 4: PolyRank WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B,C,D) -> f19(A,999,C,1) True (1,1) 1. f19(A,B,C,D) -> f19(A,-1 + B,C,D) [1 + -1*D >= 0 (?,1) && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && B >= 0] 2. f19(A,B,C,D) -> f28(A,B,999,D) [1 + -1*D >= 0 (1,1) && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && 0 >= 1 + B] 3. f28(A,B,C,D) -> f28(A,B,-1 + C,D) [1 + -1*D >= 0 (1000,1) && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && C >= 0] 4. f28(A,B,C,D) -> f36(A,B,C,D) [1 + -1*D >= 0 (1,1) && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && 0 >= 1 + C] Signature: {(f0,4);(f19,4);(f28,4);(f36,4)} Flow Graph: [0->{1},1->{1,2},2->{3},3->{3,4},4->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = 1000 p(f19) = 1 + x2 p(f28) = 1 + x2 p(f36) = 1 + x2 Following rules are strictly oriented: [1 + -1*D >= 0 ==> && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && B >= 0] f19(A,B,C,D) = 1 + B > B = f19(A,-1 + B,C,D) Following rules are weakly oriented: True ==> f0(A,B,C,D) = 1000 >= 1000 = f19(A,999,C,1) [1 + -1*D >= 0 ==> && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && 0 >= 1 + B] f19(A,B,C,D) = 1 + B >= 1 + B = f28(A,B,999,D) [1 + -1*D >= 0 ==> && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && C >= 0] f28(A,B,C,D) = 1 + B >= 1 + B = f28(A,B,-1 + C,D) [1 + -1*D >= 0 ==> && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && 0 >= 1 + C] f28(A,B,C,D) = 1 + B >= 1 + B = f36(A,B,C,D) * Step 5: KnowledgePropagation WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B,C,D) -> f19(A,999,C,1) True (1,1) 1. f19(A,B,C,D) -> f19(A,-1 + B,C,D) [1 + -1*D >= 0 (1000,1) && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && B >= 0] 2. f19(A,B,C,D) -> f28(A,B,999,D) [1 + -1*D >= 0 (1,1) && 1000 + -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*B + D >= 0 && 999 + -1*B >= 0 && 0 >= 1 + B] 3. f28(A,B,C,D) -> f28(A,B,-1 + C,D) [1 + -1*D >= 0 (1000,1) && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && C >= 0] 4. f28(A,B,C,D) -> f36(A,B,C,D) [1 + -1*D >= 0 (1,1) && 1000 + -1*C + -1*D >= 0 && -1*B + -1*D >= 0 && -1 + D >= 0 && 998 + -1*C + D >= 0 && -2 + -1*B + D >= 0 && 999 + -1*C >= 0 && 998 + -1*B + -1*C >= 0 && -1 + -1*B >= 0 && 0 >= 1 + C] Signature: {(f0,4);(f19,4);(f28,4);(f36,4)} Flow Graph: [0->{1},1->{1,2},2->{3},3->{3,4},4->{}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(1))