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(x, a()) -> x
  , f(x, g(y)) -> f(g(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: { f(x, a()) -> 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) = [3] x1 + [2] x2 + [0]
                                       
            [a] = [2]                  
                                       
        [g](x1) = [1] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
     [f(x, a())] =  [3] x + [4]        
                 >  [1] x + [0]        
                 =  [x]                
                                       
    [f(x, g(y))] =  [3] x + [2] y + [0]
                 >= [3] x + [2] y + [0]
                 =  [f(g(x), 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: { f(x, g(y)) -> f(g(x), y) }
Weak Trs: { f(x, a()) -> 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(x, g(y)) -> f(g(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).
  
    [f](x1, x2) = [2] x1 + [3] x2 + [0]
                                       
            [a] = [1]                  
                                       
        [g](x1) = [1] x1 + [2]         
  
  This order satisfies the following ordering constraints:
  
     [f(x, a())] = [2] x + [3]        
                 > [1] x + [0]        
                 = [x]                
                                      
    [f(x, g(y))] = [2] x + [3] y + [6]
                 > [2] x + [3] y + [4]
                 = [f(g(x), 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:
  { f(x, a()) -> x
  , f(x, g(y)) -> f(g(x), 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))