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(s(X), Y) -> h(s(f(h(Y), X))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add following weak dependency pairs:

Strict DPs: { f^#(s(X), Y) -> c_1(f^#(h(Y), 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^#(s(X), Y) -> c_1(f^#(h(Y), X)) }
Strict Trs: { f(s(X), Y) -> h(s(f(h(Y), 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: { f^#(s(X), Y) -> c_1(f^#(h(Y), 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:
  none

TcT has computed following constructor-restricted matrix
interpretation.

        [s](x1) = [1] x1 + [2]         
                                       
        [h](x1) = [1] x1 + [2]         
                                       
  [f^#](x1, x2) = [2] x1 + [1] x2 + [0]
                                       
      [c_1](x1) = [1]                  

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),O(1)).

Weak DPs: { f^#(s(X), Y) -> c_1(f^#(h(Y), 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.

{ f^#(s(X), Y) -> c_1(f^#(h(Y), 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))