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(x, s(y)) -> g(f(x, y), 0())
  , g(0(), f(x, x)) -> x
  , g(f(x, y), 0()) -> f(g(x, 0()), g(y, 0()))
  , g(s(x), y) -> g(f(x, y), 0()) }
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(x, s(y)) -> g(f(x, y), 0())
  , g(s(x), y) -> g(f(x, y), 0()) }

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) = [2] x1 + [2] x2 + [0]
                                       
            [0] = [0]                  
                                       
    [f](x1, x2) = [1] x1 + [1] x2 + [0]
                                       
        [s](x1) = [1] x1 + [2]         
  
  This order satisfies the following ordering constraints:
  
         [g(x, s(y))] =  [2] x + [2] y + [4]      
                      >  [2] x + [2] y + [0]      
                      =  [g(f(x, y), 0())]        
                                                  
    [g(0(), f(x, x))] =  [4] x + [0]              
                      >= [1] x + [0]              
                      =  [x]                      
                                                  
    [g(f(x, y), 0())] =  [2] x + [2] y + [0]      
                      >= [2] x + [2] y + [0]      
                      =  [f(g(x, 0()), g(y, 0()))]
                                                  
         [g(s(x), y)] =  [2] x + [2] y + [4]      
                      >  [2] x + [2] y + [0]      
                      =  [g(f(x, y), 0())]        
                                                  

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:
  { g(0(), f(x, x)) -> x
  , g(f(x, y), 0()) -> f(g(x, 0()), g(y, 0())) }
Weak Trs:
  { g(x, s(y)) -> g(f(x, y), 0())
  , g(s(x), y) -> g(f(x, y), 0()) }
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(0(), f(x, x)) -> x
  , g(f(x, y), 0()) -> f(g(x, 0()), g(y, 0())) }

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) = [2] x1 + [2] x2 + [1]
                                       
            [0] = [0]                  
                                       
    [f](x1, x2) = [1] x1 + [1] x2 + [2]
                                       
        [s](x1) = [1] x1 + [2]         
  
  This order satisfies the following ordering constraints:
  
         [g(x, s(y))] =  [2] x + [2] y + [5]      
                      >= [2] x + [2] y + [5]      
                      =  [g(f(x, y), 0())]        
                                                  
    [g(0(), f(x, x))] =  [4] x + [5]              
                      >  [1] x + [0]              
                      =  [x]                      
                                                  
    [g(f(x, y), 0())] =  [2] x + [2] y + [5]      
                      >  [2] x + [2] y + [4]      
                      =  [f(g(x, 0()), g(y, 0()))]
                                                  
         [g(s(x), y)] =  [2] x + [2] y + [5]      
                      >= [2] x + [2] y + [5]      
                      =  [g(f(x, y), 0())]        
                                                  

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(x, s(y)) -> g(f(x, y), 0())
  , g(0(), f(x, x)) -> x
  , g(f(x, y), 0()) -> f(g(x, 0()), g(y, 0()))
  , g(s(x), y) -> g(f(x, y), 0()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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