We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict Trs: { fst(0(), Z) -> nil() , fst(s(), cons(Y)) -> cons(Y) , from(X) -> cons(X) , add(0(), X) -> X , add(s(), Y) -> s() , len(nil()) -> 0() , len(cons(X)) -> s() } Obligation: runtime complexity Answer: YES(O(1),O(1)) The input is overlay and right-linear. Switching to innermost rewriting. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict Trs: { fst(0(), Z) -> nil() , fst(s(), cons(Y)) -> cons(Y) , from(X) -> cons(X) , add(0(), X) -> X , add(s(), Y) -> s() , len(nil()) -> 0() , len(cons(X)) -> s() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We add the following weak dependency pairs: Strict DPs: { fst^#(0(), Z) -> c_1() , fst^#(s(), cons(Y)) -> c_2() , from^#(X) -> c_3() , add^#(0(), X) -> c_4() , add^#(s(), Y) -> c_5() , len^#(nil()) -> c_6() , len^#(cons(X)) -> 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(1)). Strict DPs: { fst^#(0(), Z) -> c_1() , fst^#(s(), cons(Y)) -> c_2() , from^#(X) -> c_3() , add^#(0(), X) -> c_4() , add^#(s(), Y) -> c_5() , len^#(nil()) -> c_6() , len^#(cons(X)) -> c_7() } Strict Trs: { fst(0(), Z) -> nil() , fst(s(), cons(Y)) -> cons(Y) , from(X) -> cons(X) , add(0(), X) -> X , add(s(), Y) -> s() , len(nil()) -> 0() , len(cons(X)) -> s() } 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: { fst^#(0(), Z) -> c_1() , fst^#(s(), cons(Y)) -> c_2() , from^#(X) -> c_3() , add^#(0(), X) -> c_4() , add^#(s(), Y) -> c_5() , len^#(nil()) -> c_6() , len^#(cons(X)) -> c_7() } 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: none TcT has computed the following constructor-restricted matrix interpretation. [0] = [1] [0] [nil] = [0] [1] [s] = [0] [0] [cons](x1) = [0] [0] [fst^#](x1, x2) = [0 0] x1 + [0] [2 0] [0] [c_1] = [0] [0] [c_2] = [0] [0] [from^#](x1) = [1] [0] [c_3] = [0] [0] [add^#](x1, x2) = [0 1] x1 + [0] [0 0] [0] [c_4] = [0] [0] [c_5] = [0] [0] [len^#](x1) = [0] [0] [c_6] = [0] [0] [c_7] = [0] [0] The order satisfies the following ordering constraints: [fst^#(0(), Z)] = [0] [2] >= [0] [0] = [c_1()] [fst^#(s(), cons(Y))] = [0] [0] >= [0] [0] = [c_2()] [from^#(X)] = [1] [0] > [0] [0] = [c_3()] [add^#(0(), X)] = [0] [0] >= [0] [0] = [c_4()] [add^#(s(), Y)] = [0] [0] >= [0] [0] = [c_5()] [len^#(nil())] = [0] [0] >= [0] [0] = [c_6()] [len^#(cons(X))] = [0] [0] >= [0] [0] = [c_7()] 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(1)). Strict DPs: { fst^#(0(), Z) -> c_1() , fst^#(s(), cons(Y)) -> c_2() , add^#(0(), X) -> c_4() , add^#(s(), Y) -> c_5() , len^#(nil()) -> c_6() , len^#(cons(X)) -> c_7() } Weak DPs: { from^#(X) -> c_3() } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We estimate the number of application of {1,2,3,4,5,6} by applications of Pre({1,2,3,4,5,6}) = {}. Here rules are labeled as follows: DPs: { 1: fst^#(0(), Z) -> c_1() , 2: fst^#(s(), cons(Y)) -> c_2() , 3: add^#(0(), X) -> c_4() , 4: add^#(s(), Y) -> c_5() , 5: len^#(nil()) -> c_6() , 6: len^#(cons(X)) -> c_7() , 7: from^#(X) -> c_3() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { fst^#(0(), Z) -> c_1() , fst^#(s(), cons(Y)) -> c_2() , from^#(X) -> c_3() , add^#(0(), X) -> c_4() , add^#(s(), Y) -> c_5() , len^#(nil()) -> c_6() , len^#(cons(X)) -> c_7() } 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. { fst^#(0(), Z) -> c_1() , fst^#(s(), cons(Y)) -> c_2() , from^#(X) -> c_3() , add^#(0(), X) -> c_4() , add^#(s(), Y) -> c_5() , len^#(nil()) -> c_6() , len^#(cons(X)) -> c_7() } 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(1))