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(0()) -> s(0())
  , f(s(0())) -> s(0())
  , f(s(s(x))) -> f(f(s(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(s(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).
  
    [f](x1) = [1] x1 + [2]
                          
        [0] = [2]         
                          
    [s](x1) = [1] x1 + [2]
  
  This order satisfies the following ordering constraints:
  
        [f(0())] =  [4]         
                 >= [4]         
                 =  [s(0())]    
                                
     [f(s(0()))] =  [6]         
                 >  [4]         
                 =  [s(0())]    
                                
    [f(s(s(x)))] =  [1] x + [6] 
                 >= [1] x + [6] 
                 =  [f(f(s(x)))]
                                

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(0()) -> s(0())
  , f(s(s(x))) -> f(f(s(x))) }
Weak Trs: { f(s(0())) -> s(0()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.

Trs: { 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) and not(IDA(1)).
  
    [f](x1) = [1 1] x1 + [0]
              [0 2]      [0]
                            
        [0] = [2]           
              [2]           
                            
    [s](x1) = [1 0] x1 + [0]
              [0 0]      [0]
  
  This order satisfies the following ordering constraints:
  
        [f(0())] =  [4]          
                    [4]          
                 >  [2]          
                    [0]          
                 =  [s(0())]     
                                 
     [f(s(0()))] =  [2]          
                    [0]          
                 >= [2]          
                    [0]          
                 =  [s(0())]     
                                 
    [f(s(s(x)))] =  [1 0] x + [0]
                    [0 0]     [0]
                 >= [1 0] x + [0]
                    [0 0]     [0]
                 =  [f(f(s(x)))] 
                                 

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(s(x))) -> f(f(s(x))) }
Weak Trs:
  { f(0()) -> s(0())
  , f(s(0())) -> s(0()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.

Trs: { f(s(s(x))) -> f(f(s(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) and not(IDA(1)).
  
    [f](x1) = [1 1] x1 + [0]
              [0 0]      [1]
                            
        [0] = [2]           
              [2]           
                            
    [s](x1) = [1 0] x1 + [2]
              [0 0]      [0]
  
  This order satisfies the following ordering constraints:
  
        [f(0())] =  [4]          
                    [1]          
                 >= [4]          
                    [0]          
                 =  [s(0())]     
                                 
     [f(s(0()))] =  [4]          
                    [1]          
                 >= [4]          
                    [0]          
                 =  [s(0())]     
                                 
    [f(s(s(x)))] =  [1 0] x + [4]
                    [0 0]     [1]
                 >  [1 0] x + [3]
                    [0 0]     [1]
                 =  [f(f(s(x)))] 
                                 

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

Empty rules are trivially bounded

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