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 use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs: { f(s(X), Y) -> h(s(f(h(Y), X))) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [f](x1, x2) = [2] x1 + [2] x2 + [0]
                                       
        [s](x1) = [1] x1 + [2]         
                                       
        [h](x1) = [1] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
    [f(s(X), Y)] = [2] X + [2] Y + [4]
                 > [2] X + [2] Y + [2]
                 = [h(s(f(h(Y), X)))] 
                                      

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).

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

Empty rules are trivially bounded

Hurray, we answered YES(O(1),O(n^1))