YES(O(1),O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { f(f(x)) -> f(c(f(x))) , f(f(x)) -> f(d(f(x))) , g(c(x)) -> x , g(c(h(0()))) -> g(d(1())) , g(c(1())) -> g(d(h(0()))) , g(d(x)) -> x , g(h(x)) -> g(x) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following innermost weak dependency pairs: Strict DPs: { f^#(f(x)) -> c_1(f^#(c(f(x)))) , f^#(f(x)) -> c_2(f^#(d(f(x)))) , g^#(c(x)) -> c_3() , g^#(c(h(0()))) -> c_4(g^#(d(1()))) , g^#(c(1())) -> c_5(g^#(d(h(0())))) , g^#(d(x)) -> c_6() , g^#(h(x)) -> c_7(g^#(x)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { f^#(f(x)) -> c_1(f^#(c(f(x)))) , f^#(f(x)) -> c_2(f^#(d(f(x)))) , g^#(c(x)) -> c_3() , g^#(c(h(0()))) -> c_4(g^#(d(1()))) , g^#(c(1())) -> c_5(g^#(d(h(0())))) , g^#(d(x)) -> c_6() , g^#(h(x)) -> c_7(g^#(x)) } Strict Trs: { f(f(x)) -> f(c(f(x))) , f(f(x)) -> f(d(f(x))) , g(c(x)) -> x , g(c(h(0()))) -> g(d(1())) , g(c(1())) -> g(d(h(0()))) , g(d(x)) -> x , g(h(x)) -> g(x) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We replace rewrite rules by usable rules: Strict Usable Rules: { f(f(x)) -> f(c(f(x))) , f(f(x)) -> f(d(f(x))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { f^#(f(x)) -> c_1(f^#(c(f(x)))) , f^#(f(x)) -> c_2(f^#(d(f(x)))) , g^#(c(x)) -> c_3() , g^#(c(h(0()))) -> c_4(g^#(d(1()))) , g^#(c(1())) -> c_5(g^#(d(h(0())))) , g^#(d(x)) -> c_6() , g^#(h(x)) -> c_7(g^#(x)) } Strict Trs: { f(f(x)) -> f(c(f(x))) , f(f(x)) -> f(d(f(x))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(c_4) = {1}, Uargs(c_5) = {1}, Uargs(c_7) = {1} TcT has computed the following constructor-restricted matrix interpretation. [f](x1) = [0 2] x1 + [0] [0 0] [1] [c](x1) = [0] [0] [d](x1) = [0] [0] [h](x1) = [0] [0] [0] = [0] [0] [1] = [0] [0] [f^#](x1) = [2] [0] [c_1](x1) = [2] [2] [c_2](x1) = [2] [2] [g^#](x1) = [0] [0] [c_3] = [1] [1] [c_4](x1) = [1 0] x1 + [2] [0 1] [2] [c_5](x1) = [1 0] x1 + [2] [0 1] [2] [c_6] = [1] [1] [c_7](x1) = [1 0] x1 + [2] [0 1] [2] The following symbols are considered usable {f, f^#, g^#} The order satisfies the following ordering constraints: [f(f(x))] = [2] [1] > [0] [1] = [f(c(f(x)))] [f(f(x))] = [2] [1] > [0] [1] = [f(d(f(x)))] [f^#(f(x))] = [2] [0] ? [2] [2] = [c_1(f^#(c(f(x))))] [f^#(f(x))] = [2] [0] ? [2] [2] = [c_2(f^#(d(f(x))))] [g^#(c(x))] = [0] [0] ? [1] [1] = [c_3()] [g^#(c(h(0())))] = [0] [0] ? [2] [2] = [c_4(g^#(d(1())))] [g^#(c(1()))] = [0] [0] ? [2] [2] = [c_5(g^#(d(h(0()))))] [g^#(d(x))] = [0] [0] ? [1] [1] = [c_6()] [g^#(h(x))] = [0] [0] ? [2] [2] = [c_7(g^#(x))] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { f^#(f(x)) -> c_1(f^#(c(f(x)))) , f^#(f(x)) -> c_2(f^#(d(f(x)))) , g^#(c(x)) -> c_3() , g^#(c(h(0()))) -> c_4(g^#(d(1()))) , g^#(c(1())) -> c_5(g^#(d(h(0())))) , g^#(d(x)) -> c_6() , g^#(h(x)) -> c_7(g^#(x)) } Weak Trs: { f(f(x)) -> f(c(f(x))) , f(f(x)) -> f(d(f(x))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Consider the dependency graph: 1: f^#(f(x)) -> c_1(f^#(c(f(x)))) 2: f^#(f(x)) -> c_2(f^#(d(f(x)))) 3: g^#(c(x)) -> c_3() 4: g^#(c(h(0()))) -> c_4(g^#(d(1()))) -->_1 g^#(d(x)) -> c_6() :6 5: g^#(c(1())) -> c_5(g^#(d(h(0())))) -->_1 g^#(d(x)) -> c_6() :6 6: g^#(d(x)) -> c_6() 7: g^#(h(x)) -> c_7(g^#(x)) -->_1 g^#(h(x)) -> c_7(g^#(x)) :7 -->_1 g^#(d(x)) -> c_6() :6 -->_1 g^#(c(1())) -> c_5(g^#(d(h(0())))) :5 -->_1 g^#(c(h(0()))) -> c_4(g^#(d(1()))) :4 -->_1 g^#(c(x)) -> c_3() :3 Only the nodes {3,4,6,5,7} are reachable from nodes {3,4,5,6,7} that start derivation from marked basic terms. The nodes not reachable are removed from the problem. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { g^#(c(x)) -> c_3() , g^#(c(h(0()))) -> c_4(g^#(d(1()))) , g^#(c(1())) -> c_5(g^#(d(h(0())))) , g^#(d(x)) -> c_6() , g^#(h(x)) -> c_7(g^#(x)) } Weak Trs: { f(f(x)) -> f(c(f(x))) , f(f(x)) -> f(d(f(x))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {1,4} by applications of Pre({1,4}) = {2,3,5}. Here rules are labeled as follows: DPs: { 1: g^#(c(x)) -> c_3() , 2: g^#(c(h(0()))) -> c_4(g^#(d(1()))) , 3: g^#(c(1())) -> c_5(g^#(d(h(0())))) , 4: g^#(d(x)) -> c_6() , 5: g^#(h(x)) -> c_7(g^#(x)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { g^#(c(h(0()))) -> c_4(g^#(d(1()))) , g^#(c(1())) -> c_5(g^#(d(h(0())))) , g^#(h(x)) -> c_7(g^#(x)) } Weak DPs: { g^#(c(x)) -> c_3() , g^#(d(x)) -> c_6() } Weak Trs: { f(f(x)) -> f(c(f(x))) , f(f(x)) -> f(d(f(x))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {1,2} by applications of Pre({1,2}) = {3}. Here rules are labeled as follows: DPs: { 1: g^#(c(h(0()))) -> c_4(g^#(d(1()))) , 2: g^#(c(1())) -> c_5(g^#(d(h(0())))) , 3: g^#(h(x)) -> c_7(g^#(x)) , 4: g^#(c(x)) -> c_3() , 5: g^#(d(x)) -> c_6() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { g^#(h(x)) -> c_7(g^#(x)) } Weak DPs: { g^#(c(x)) -> c_3() , g^#(c(h(0()))) -> c_4(g^#(d(1()))) , g^#(c(1())) -> c_5(g^#(d(h(0())))) , g^#(d(x)) -> c_6() } Weak Trs: { f(f(x)) -> f(c(f(x))) , f(f(x)) -> f(d(f(x))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { g^#(c(x)) -> c_3() , g^#(c(h(0()))) -> c_4(g^#(d(1()))) , g^#(c(1())) -> c_5(g^#(d(h(0())))) , g^#(d(x)) -> c_6() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { g^#(h(x)) -> c_7(g^#(x)) } Weak Trs: { f(f(x)) -> f(c(f(x))) , f(f(x)) -> f(d(f(x))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^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(n^1)). Strict DPs: { g^#(h(x)) -> c_7(g^#(x)) } 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: g^#(h(x)) -> c_7(g^#(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(f) = {}, safe(c) = {1}, safe(d) = {1}, safe(g) = {}, safe(h) = {1}, safe(0) = {}, safe(1) = {}, safe(f^#) = {}, safe(c_1) = {}, safe(c_2) = {}, safe(g^#) = {}, safe(c_3) = {}, safe(c_4) = {}, safe(c_5) = {}, safe(c_6) = {}, safe(c_7) = {} and precedence empty . Following symbols are considered recursive: {g^#} The recursion depth is 1. Further, following argument filtering is employed: pi(f) = [], pi(c) = [], pi(d) = [], pi(g) = [], pi(h) = [1], pi(0) = [], pi(1) = [], pi(f^#) = [], pi(c_1) = [], pi(c_2) = [], pi(g^#) = [1], pi(c_3) = [], pi(c_4) = [], pi(c_5) = [], pi(c_6) = [], pi(c_7) = [1] Usable defined function symbols are a subset of: {f^#, g^#} For your convenience, here are the satisfied ordering constraints: pi(g^#(h(x))) = g^#(h(; x);) > c_7(g^#(x;);) = pi(c_7(g^#(x))) 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: { g^#(h(x)) -> c_7(g^#(x)) } 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. { g^#(h(x)) -> c_7(g^#(x)) } 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^1))