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:
  { g(f(x), y) -> f(h(x, y))
  , h(x, y) -> g(x, f(y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(f) = {1}

TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).

  [g](x1, x2) = [1] x1 + [1] x2 + [0]
                                     
      [f](x1) = [1] x1 + [0]         
                                     
  [h](x1, x2) = [1] x1 + [1] x2 + [1]

The following symbols are considered usable

  {g, h}

The order satisfies the following ordering constraints:

  [g(f(x), y)] = [1] x + [1] y + [0]
               ? [1] x + [1] y + [1]
               = [f(h(x, y))]       
                                    
     [h(x, y)] = [1] x + [1] y + [1]
               > [1] x + [1] y + [0]
               = [g(x, f(y))]       
                                    

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(n^1)).

Strict Trs: { g(f(x), y) -> f(h(x, y)) }
Weak Trs: { h(x, y) -> g(x, f(y)) }
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: { g(f(x), y) -> f(h(x, y)) }

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:
----------
  The following argument positions are usable:
    Uargs(f) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [g](x1, x2) = [2] x1 + [4]
                              
        [f](x1) = [1] x1 + [4]
                              
    [h](x1, x2) = [2] x1 + [4]
  
  The following symbols are considered usable
  
    {g, h}
  
  The order satisfies the following ordering constraints:
  
    [g(f(x), y)] =  [2] x + [12]
                 >  [2] x + [8] 
                 =  [f(h(x, y))]
                                
       [h(x, y)] =  [2] x + [4] 
                 >= [2] x + [4] 
                 =  [g(x, f(y))]
                                

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:
  { g(f(x), y) -> f(h(x, y))
  , h(x, y) -> g(x, f(y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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