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))

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [g](x1, x2) = [3] x1 + [1] x2 + [0]
                                       
        [f](x1) = [1] x1 + [2]         
                                       
    [h](x1, x2) = [3] x1 + [1] x2 + [2]
  
  This order satisfies the following ordering constraints:
  
    [g(f(x), y)] =  [3] x + [1] y + [6]
                 >  [3] x + [1] y + [4]
                 =  [f(h(x, y))]       
                                       
       [h(x, y)] =  [3] x + [1] y + [2]
                 >= [3] x + [1] y + [2]
                 =  [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(n^1)).

Strict Trs: { h(x, y) -> g(x, f(y)) }
Weak Trs: { g(f(x), y) -> f(h(x, 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: { h(x, y) -> g(x, f(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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [g](x1, x2) = [3] x1 + [1] x2 + [0]
                                       
        [f](x1) = [1] x1 + [2]         
                                       
    [h](x1, x2) = [3] x1 + [1] x2 + [3]
  
  This order satisfies the following ordering constraints:
  
    [g(f(x), y)] = [3] x + [1] y + [6]
                 > [3] x + [1] y + [5]
                 = [f(h(x, y))]       
                                      
       [h(x, y)] = [3] x + [1] y + [3]
                 > [3] x + [1] y + [2]
                 = [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))