YES(O(1),O(n^3)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^3)). Strict Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() , log(s(s(x))) -> s(log(s(quot(x, s(s(0())))))) , log(s(0())) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^3)) We add the following dependency tuples: Strict DPs: { pred^#(s(x)) -> c_1() , minus^#(x, s(y)) -> c_2(pred^#(minus(x, y)), minus^#(x, y)) , minus^#(x, 0()) -> c_3() , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y)), minus^#(x, y)) , quot^#(0(), s(y)) -> c_5() , log^#(s(s(x))) -> c_6(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) , log^#(s(0())) -> c_7() } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^3)). Strict DPs: { pred^#(s(x)) -> c_1() , minus^#(x, s(y)) -> c_2(pred^#(minus(x, y)), minus^#(x, y)) , minus^#(x, 0()) -> c_3() , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y)), minus^#(x, y)) , quot^#(0(), s(y)) -> c_5() , log^#(s(s(x))) -> c_6(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) , log^#(s(0())) -> c_7() } Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() , log(s(s(x))) -> s(log(s(quot(x, s(s(0())))))) , log(s(0())) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^3)) We estimate the number of application of {1,3,5,7} by applications of Pre({1,3,5,7}) = {2,4,6}. Here rules are labeled as follows: DPs: { 1: pred^#(s(x)) -> c_1() , 2: minus^#(x, s(y)) -> c_2(pred^#(minus(x, y)), minus^#(x, y)) , 3: minus^#(x, 0()) -> c_3() , 4: quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y)), minus^#(x, y)) , 5: quot^#(0(), s(y)) -> c_5() , 6: log^#(s(s(x))) -> c_6(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) , 7: log^#(s(0())) -> c_7() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^3)). Strict DPs: { minus^#(x, s(y)) -> c_2(pred^#(minus(x, y)), minus^#(x, y)) , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y)), minus^#(x, y)) , log^#(s(s(x))) -> c_6(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) } Weak DPs: { pred^#(s(x)) -> c_1() , minus^#(x, 0()) -> c_3() , quot^#(0(), s(y)) -> c_5() , log^#(s(0())) -> c_7() } Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() , log(s(s(x))) -> s(log(s(quot(x, s(s(0())))))) , log(s(0())) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^3)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { pred^#(s(x)) -> c_1() , minus^#(x, 0()) -> c_3() , quot^#(0(), s(y)) -> c_5() , log^#(s(0())) -> c_7() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^3)). Strict DPs: { minus^#(x, s(y)) -> c_2(pred^#(minus(x, y)), minus^#(x, y)) , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y)), minus^#(x, y)) , log^#(s(s(x))) -> c_6(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) } Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() , log(s(s(x))) -> s(log(s(quot(x, s(s(0())))))) , log(s(0())) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^3)) Due to missing edges in the dependency-graph, the right-hand sides of following rules could be simplified: { minus^#(x, s(y)) -> c_2(pred^#(minus(x, y)), minus^#(x, y)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^3)). Strict DPs: { minus^#(x, s(y)) -> c_1(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) , log^#(s(s(x))) -> c_3(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) } Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() , log(s(s(x))) -> s(log(s(quot(x, s(s(0())))))) , log(s(0())) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^3)) We replace rewrite rules by usable rules: Weak Usable Rules: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^3)). Strict DPs: { minus^#(x, s(y)) -> c_1(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) , log^#(s(s(x))) -> c_3(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) } Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^3)) We decompose the input problem according to the dependency graph into the upper component { log^#(s(s(x))) -> c_3(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) } and lower component { minus^#(x, s(y)) -> c_1(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) } Further, following extension rules are added to the lower component. { log^#(s(s(x))) -> quot^#(x, s(s(0()))) , log^#(s(s(x))) -> log^#(s(quot(x, s(s(0()))))) } TcT solves the upper component with certificate YES(O(1),O(n^1)). Sub-proof: ---------- We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { log^#(s(s(x))) -> c_3(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) } Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'Small Polynomial Path Order (PS,1-bounded)' to orient following rules strictly. DPs: { 1: log^#(s(s(x))) -> c_3(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) } Trs: { pred(s(x)) -> x } Sub-proof: ---------- The input was oriented with the instance of 'Small Polynomial Path Order (PS,1-bounded)' as induced by the safe mapping safe(pred) = {1}, safe(s) = {1}, safe(minus) = {1, 2}, safe(0) = {}, safe(quot) = {1, 2}, safe(log) = {}, safe(pred^#) = {}, safe(c_1) = {}, safe(minus^#) = {}, safe(c_2) = {}, safe(c_3) = {}, safe(quot^#) = {}, safe(c_4) = {}, safe(c_5) = {}, safe(log^#) = {}, safe(c_6) = {}, safe(c_7) = {}, safe(c) = {}, safe(c_1) = {}, safe(c_2) = {}, safe(c_3) = {} and precedence minus > pred, quot > pred, log^# > pred, minus ~ quot, minus ~ log^#, quot ~ log^# . Following symbols are considered recursive: {minus, quot, log^#} The recursion depth is 1. Further, following argument filtering is employed: pi(pred) = 1, pi(s) = [1], pi(minus) = 1, pi(0) = [], pi(quot) = 1, pi(log) = [], pi(pred^#) = [], pi(c_1) = [], pi(minus^#) = [], pi(c_2) = [], pi(c_3) = [], pi(quot^#) = [1], pi(c_4) = [], pi(c_5) = [], pi(log^#) = [1], pi(c_6) = [], pi(c_7) = [], pi(c) = [], pi(c_1) = [], pi(c_2) = [], pi(c_3) = [1, 2] Usable defined function symbols are a subset of: {pred, minus, quot, pred^#, minus^#, quot^#, log^#} For your convenience, here are the satisfied ordering constraints: pi(log^#(s(s(x)))) = log^#(s(; s(; x));) > c_3(log^#(s(; x);), quot^#(x;);) = pi(c_3(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0()))))) pi(pred(s(x))) = s(; x) > x = pi(x) pi(minus(x, s(y))) = x >= x = pi(pred(minus(x, y))) pi(minus(x, 0())) = x >= x = pi(x) pi(quot(s(x), s(y))) = s(; x) >= s(; x) = pi(s(quot(minus(x, y), s(y)))) pi(quot(0(), s(y))) = 0() >= 0() = pi(0()) The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { log^#(s(s(x))) -> c_3(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) } Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { log^#(s(s(x))) -> c_3(log^#(s(quot(x, s(s(0()))))), quot^#(x, s(s(0())))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) No rule is usable, rules are removed from the input problem. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Rules: Empty Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict DPs: { minus^#(x, s(y)) -> c_1(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) } Weak DPs: { log^#(s(s(x))) -> quot^#(x, s(s(0()))) , log^#(s(s(x))) -> log^#(s(quot(x, s(s(0()))))) } Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We decompose the input problem according to the dependency graph into the upper component { quot^#(s(x), s(y)) -> c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) , log^#(s(s(x))) -> quot^#(x, s(s(0()))) , log^#(s(s(x))) -> log^#(s(quot(x, s(s(0()))))) } and lower component { minus^#(x, s(y)) -> c_1(minus^#(x, y)) } Further, following extension rules are added to the lower component. { quot^#(s(x), s(y)) -> minus^#(x, y) , quot^#(s(x), s(y)) -> quot^#(minus(x, y), s(y)) , log^#(s(s(x))) -> quot^#(x, s(s(0()))) , log^#(s(s(x))) -> log^#(s(quot(x, s(s(0()))))) } TcT solves the upper component with certificate YES(O(1),O(n^1)). Sub-proof: ---------- We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { quot^#(s(x), s(y)) -> c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) } Weak DPs: { log^#(s(s(x))) -> quot^#(x, s(s(0()))) , log^#(s(s(x))) -> log^#(s(quot(x, s(s(0()))))) } Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'Small Polynomial Path Order (PS,1-bounded)' to orient following rules strictly. DPs: { 1: quot^#(s(x), s(y)) -> c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) , 2: log^#(s(s(x))) -> quot^#(x, s(s(0()))) , 3: log^#(s(s(x))) -> log^#(s(quot(x, s(s(0()))))) } Trs: { pred(s(x)) -> x } Sub-proof: ---------- The input was oriented with the instance of 'Small Polynomial Path Order (PS,1-bounded)' as induced by the safe mapping safe(pred) = {}, safe(s) = {1}, safe(minus) = {}, safe(0) = {}, safe(quot) = {}, safe(log) = {}, safe(pred^#) = {}, safe(c_1) = {}, safe(minus^#) = {}, safe(c_2) = {}, safe(c_3) = {}, safe(quot^#) = {2}, safe(c_4) = {}, safe(c_5) = {}, safe(log^#) = {}, safe(c_6) = {}, safe(c_7) = {}, safe(c) = {}, safe(c_1) = {}, safe(c_2) = {}, safe(c_3) = {} and precedence pred ~ minus, pred ~ quot, minus ~ quot, quot^# ~ log^# . Following symbols are considered recursive: {quot^#, log^#} The recursion depth is 1. Further, following argument filtering is employed: pi(pred) = 1, pi(s) = [1], pi(minus) = 1, pi(0) = [], pi(quot) = 1, pi(log) = [], pi(pred^#) = [], pi(c_1) = [], pi(minus^#) = [], pi(c_2) = [], pi(c_3) = [], pi(quot^#) = [1], pi(c_4) = [], pi(c_5) = [], pi(log^#) = [1], pi(c_6) = [], pi(c_7) = [], pi(c) = [], pi(c_1) = [], pi(c_2) = [1, 2], pi(c_3) = [] Usable defined function symbols are a subset of: {pred, minus, quot, pred^#, minus^#, quot^#, log^#} For your convenience, here are the satisfied ordering constraints: pi(quot^#(s(x), s(y))) = quot^#(s(; x);) > c_2(quot^#(x;), minus^#();) = pi(c_2(quot^#(minus(x, y), s(y)), minus^#(x, y))) pi(log^#(s(s(x)))) = log^#(s(; s(; x));) > quot^#(x;) = pi(quot^#(x, s(s(0())))) pi(log^#(s(s(x)))) = log^#(s(; s(; x));) > log^#(s(; x);) = pi(log^#(s(quot(x, s(s(0())))))) pi(pred(s(x))) = s(; x) > x = pi(x) pi(minus(x, s(y))) = x >= x = pi(pred(minus(x, y))) pi(minus(x, 0())) = x >= x = pi(x) pi(quot(s(x), s(y))) = s(; x) >= s(; x) = pi(s(quot(minus(x, y), s(y)))) pi(quot(0(), s(y))) = 0() >= 0() = pi(0()) The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { quot^#(s(x), s(y)) -> c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) , log^#(s(s(x))) -> quot^#(x, s(s(0()))) , log^#(s(s(x))) -> log^#(s(quot(x, s(s(0()))))) } Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { quot^#(s(x), s(y)) -> c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) , log^#(s(s(x))) -> quot^#(x, s(s(0()))) , log^#(s(s(x))) -> log^#(s(quot(x, s(s(0()))))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) No rule is usable, rules are removed from the input problem. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Rules: Empty Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { minus^#(x, s(y)) -> c_1(minus^#(x, y)) } Weak DPs: { quot^#(s(x), s(y)) -> minus^#(x, y) , quot^#(s(x), s(y)) -> quot^#(minus(x, y), s(y)) , log^#(s(s(x))) -> quot^#(x, s(s(0()))) , log^#(s(s(x))) -> log^#(s(quot(x, s(s(0()))))) } Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 1: minus^#(x, s(y)) -> c_1(minus^#(x, y)) , 2: quot^#(s(x), s(y)) -> minus^#(x, y) , 4: log^#(s(s(x))) -> quot^#(x, s(s(0()))) } Trs: { pred(s(x)) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_1) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [pred](x1) = [1] x1 + [0] [s](x1) = [1] x1 + [1] [minus](x1, x2) = [1] x1 + [0] [0] = [0] [quot](x1, x2) = [4] x1 + [0] [log](x1) = [7] x1 + [0] [pred^#](x1) = [7] x1 + [0] [c_1] = [0] [minus^#](x1, x2) = [1] x2 + [0] [c_2](x1, x2) = [7] x1 + [7] x2 + [0] [c_3] = [0] [quot^#](x1, x2) = [1] x2 + [0] [c_4](x1, x2) = [7] x1 + [7] x2 + [0] [c_5] = [0] [log^#](x1) = [4] [c_6](x1, x2) = [7] x1 + [7] x2 + [0] [c_7] = [0] [c] = [0] [c_1](x1) = [1] x1 + [0] [c_2](x1, x2) = [7] x1 + [7] x2 + [0] [c_3](x1, x2) = [7] x1 + [7] x2 + [0] The following symbols are considered usable {pred, minus, quot, minus^#, quot^#, log^#} The order satisfies the following ordering constraints: [pred(s(x))] = [1] x + [1] > [1] x + [0] = [x] [minus(x, s(y))] = [1] x + [0] >= [1] x + [0] = [pred(minus(x, y))] [minus(x, 0())] = [1] x + [0] >= [1] x + [0] = [x] [quot(s(x), s(y))] = [4] x + [4] > [4] x + [1] = [s(quot(minus(x, y), s(y)))] [quot(0(), s(y))] = [0] >= [0] = [0()] [minus^#(x, s(y))] = [1] y + [1] > [1] y + [0] = [c_1(minus^#(x, y))] [quot^#(s(x), s(y))] = [1] y + [1] > [1] y + [0] = [minus^#(x, y)] [quot^#(s(x), s(y))] = [1] y + [1] >= [1] y + [1] = [quot^#(minus(x, y), s(y))] [log^#(s(s(x)))] = [4] > [2] = [quot^#(x, s(s(0())))] [log^#(s(s(x)))] = [4] >= [4] = [log^#(s(quot(x, s(s(0())))))] The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { minus^#(x, s(y)) -> c_1(minus^#(x, y)) , quot^#(s(x), s(y)) -> minus^#(x, y) , quot^#(s(x), s(y)) -> quot^#(minus(x, y), s(y)) , log^#(s(s(x))) -> quot^#(x, s(s(0()))) , log^#(s(s(x))) -> log^#(s(quot(x, s(s(0()))))) } Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { minus^#(x, s(y)) -> c_1(minus^#(x, y)) , quot^#(s(x), s(y)) -> minus^#(x, y) , quot^#(s(x), s(y)) -> quot^#(minus(x, y), s(y)) , log^#(s(s(x))) -> quot^#(x, s(s(0()))) , log^#(s(s(x))) -> log^#(s(quot(x, s(s(0()))))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { pred(s(x)) -> x , minus(x, s(y)) -> pred(minus(x, y)) , minus(x, 0()) -> x , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) , quot(0(), s(y)) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) No rule is usable, rules are removed from the input problem. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Rules: Empty Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^3))