YES(O(1),O(1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict Trs: { terms(N) -> cons(recip(sqr(N))) , sqr(0()) -> 0() , sqr(s()) -> s() , dbl(0()) -> 0() , dbl(s()) -> s() , add(0(), X) -> X , add(s(), Y) -> s() , first(0(), X) -> nil() , first(s(), cons(Y)) -> cons(Y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We add following weak dependency pairs: Strict DPs: { terms^#(N) -> c_1(sqr^#(N)) , sqr^#(0()) -> c_2() , sqr^#(s()) -> c_3() , dbl^#(0()) -> c_4() , dbl^#(s()) -> c_5() , add^#(0(), X) -> c_6() , add^#(s(), Y) -> c_7() , first^#(0(), X) -> c_8() , first^#(s(), cons(Y)) -> c_9() } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { terms^#(N) -> c_1(sqr^#(N)) , sqr^#(0()) -> c_2() , sqr^#(s()) -> c_3() , dbl^#(0()) -> c_4() , dbl^#(s()) -> c_5() , add^#(0(), X) -> c_6() , add^#(s(), Y) -> c_7() , first^#(0(), X) -> c_8() , first^#(s(), cons(Y)) -> c_9() } Strict Trs: { terms(N) -> cons(recip(sqr(N))) , sqr(0()) -> 0() , sqr(s()) -> s() , dbl(0()) -> 0() , dbl(s()) -> s() , add(0(), X) -> X , add(s(), Y) -> s() , first(0(), X) -> nil() , first(s(), cons(Y)) -> cons(Y) } 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)). Strict DPs: { terms^#(N) -> c_1(sqr^#(N)) , sqr^#(0()) -> c_2() , sqr^#(s()) -> c_3() , dbl^#(0()) -> c_4() , dbl^#(s()) -> c_5() , add^#(0(), X) -> c_6() , add^#(s(), Y) -> c_7() , first^#(0(), X) -> c_8() , first^#(s(), cons(Y)) -> c_9() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(c_1) = {1} TcT has computed following constructor-restricted matrix interpretation. [cons](x1) = [2] [0] = [2] [s] = [2] [terms^#](x1) = [2] x1 + [2] [c_1](x1) = [1] x1 + [1] [sqr^#](x1) = [2] x1 + [2] [c_2] = [1] [c_3] = [1] [dbl^#](x1) = [2] x1 + [2] [c_4] = [1] [c_5] = [1] [add^#](x1, x2) = [2] x1 + [2] [c_6] = [1] [c_7] = [1] [first^#](x1, x2) = [2] x1 + [1] x2 + [1] [c_8] = [0] [c_9] = [2] This order satisfies following ordering constraints: 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)). Strict DPs: { terms^#(N) -> c_1(sqr^#(N)) } Weak DPs: { sqr^#(0()) -> c_2() , sqr^#(s()) -> c_3() , dbl^#(0()) -> c_4() , dbl^#(s()) -> c_5() , add^#(0(), X) -> c_6() , add^#(s(), Y) -> c_7() , first^#(0(), X) -> c_8() , first^#(s(), cons(Y)) -> c_9() } Obligation: innermost runtime complexity Answer: YES(?,O(1)) We estimate the number of application of {1} by applications of Pre({1}) = {}. Here rules are labeled as follows: DPs: { 1: terms^#(N) -> c_1(sqr^#(N)) , 2: sqr^#(0()) -> c_2() , 3: sqr^#(s()) -> c_3() , 4: dbl^#(0()) -> c_4() , 5: dbl^#(s()) -> c_5() , 6: add^#(0(), X) -> c_6() , 7: add^#(s(), Y) -> c_7() , 8: first^#(0(), X) -> c_8() , 9: first^#(s(), cons(Y)) -> c_9() } We are left with following problem, upon which TcT provides the certificate YES(?,O(1)). Weak DPs: { terms^#(N) -> c_1(sqr^#(N)) , sqr^#(0()) -> c_2() , sqr^#(s()) -> c_3() , dbl^#(0()) -> c_4() , dbl^#(s()) -> c_5() , add^#(0(), X) -> c_6() , add^#(s(), Y) -> c_7() , first^#(0(), X) -> c_8() , first^#(s(), cons(Y)) -> c_9() } Obligation: innermost runtime complexity Answer: YES(?,O(1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { terms^#(N) -> c_1(sqr^#(N)) , sqr^#(0()) -> c_2() , sqr^#(s()) -> c_3() , dbl^#(0()) -> c_4() , dbl^#(s()) -> c_5() , add^#(0(), X) -> c_6() , add^#(s(), Y) -> c_7() , first^#(0(), X) -> c_8() , first^#(s(), cons(Y)) -> c_9() } We are left with following problem, upon which TcT provides the certificate YES(?,O(1)). Rules: Empty Obligation: innermost runtime complexity Answer: YES(?,O(1)) The input was oriented with the instance of 'Small Polynomial Path Order (PS)' as induced by the safe mapping and precedence empty . Following symbols are considered recursive: {} The recursion depth is 0. Further, following argument filtering is employed: empty Usable defined function symbols are a subset of: {} For your convenience, here are the satisfied ordering constraints: Hurray, we answered YES(O(1),O(1))