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(s(x)) -> f(x)
  , g(0()) -> 0()
  , f(s(x)) -> s(s(g(x)))
  , f(0()) -> s(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()) -> 0()
  , f(0()) -> s(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) = [2] x1 + [0]
                          
    [s](x1) = [1] x1 + [0]
                          
    [f](x1) = [2] x1 + [0]
                          
        [0] = [2]         
  
  This order satisfies the following ordering constraints:
  
    [g(s(x))] =  [2] x + [0] 
              >= [2] x + [0] 
              =  [f(x)]      
                             
     [g(0())] =  [4]         
              >  [2]         
              =  [0()]       
                             
    [f(s(x))] =  [2] x + [0] 
              >= [2] x + [0] 
              =  [s(s(g(x)))]
                             
     [f(0())] =  [4]         
              >  [2]         
              =  [s(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(s(x)) -> f(x)
  , f(s(x)) -> s(s(g(x))) }
Weak Trs:
  { g(0()) -> 0()
  , f(0()) -> s(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(s(x)) -> f(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).
  
    [g](x1) = [2] x1 + [0]
                          
    [s](x1) = [1] x1 + [2]
                          
    [f](x1) = [2] x1 + [0]
                          
        [0] = [2]         
  
  This order satisfies the following ordering constraints:
  
    [g(s(x))] =  [2] x + [4] 
              >  [2] x + [0] 
              =  [f(x)]      
                             
     [g(0())] =  [4]         
              >  [2]         
              =  [0()]       
                             
    [f(s(x))] =  [2] x + [4] 
              >= [2] x + [4] 
              =  [s(s(g(x)))]
                             
     [f(0())] =  [4]         
              >= [4]         
              =  [s(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: { f(s(x)) -> s(s(g(x))) }
Weak Trs:
  { g(s(x)) -> f(x)
  , g(0()) -> 0()
  , f(0()) -> s(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: { f(s(x)) -> s(s(g(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).
  
    [g](x1) = [2] x1 + [0]
                          
    [s](x1) = [1] x1 + [2]
                          
    [f](x1) = [2] x1 + [1]
                          
        [0] = [2]         
  
  This order satisfies the following ordering constraints:
  
    [g(s(x))] = [2] x + [4] 
              > [2] x + [1] 
              = [f(x)]      
                            
     [g(0())] = [4]         
              > [2]         
              = [0()]       
                            
    [f(s(x))] = [2] x + [5] 
              > [2] x + [4] 
              = [s(s(g(x)))]
                            
     [f(0())] = [5]         
              > [4]         
              = [s(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(s(x)) -> f(x)
  , g(0()) -> 0()
  , f(s(x)) -> s(s(g(x)))
  , f(0()) -> s(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))