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: { add(0(), x) -> x , add(s(x), y) -> s(add(x, y)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add following weak dependency pairs: Strict DPs: { add^#(0(), x) -> c_1() , add^#(s(x), y) -> c_2(add^#(x, y)) } 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: { add^#(0(), x) -> c_1() , add^#(s(x), y) -> c_2(add^#(x, y)) } Strict Trs: { add(0(), x) -> x , add(s(x), y) -> s(add(x, y)) } 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: { add^#(0(), x) -> c_1() , add^#(s(x), y) -> c_2(add^#(x, y)) } 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_2) = {1} TcT has computed following constructor-restricted matrix interpretation. [0] = [2] [s](x1) = [1] x1 + [1] [add^#](x1, x2) = [2] x2 + [1] [c_1] = [0] [c_2](x1) = [1] x1 + [0] 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(n^1)). Strict DPs: { add^#(s(x), y) -> c_2(add^#(x, y)) } Weak DPs: { add^#(0(), x) -> c_1() } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { add^#(0(), x) -> c_1() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { add^#(s(x), y) -> c_2(add^#(x, y)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) The input was oriented with the instance of 'Small Polynomial Path Order (PS)' as induced by the safe mapping safe(s) = {1}, safe(add^#) = {2}, safe(c_2) = {} and precedence empty . Following symbols are considered recursive: {add^#} The recursion depth is 1. Further, following argument filtering is employed: pi(s) = [1], pi(add^#) = [1, 2], pi(c_2) = [1] Usable defined function symbols are a subset of: {add^#} For your convenience, here are the satisfied ordering constraints: pi(add^#(s(x), y)) = add^#(s(; x); y) > c_2(add^#(x; y);) = pi(c_2(add^#(x, y))) Hurray, we answered YES(O(1),O(n^1))