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, y, s(z)) -> s(f(0(), 1(), z))
  , f(0(), 1(), x) -> f(s(x), x, x) }
Obligation:
  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, y, s(z)) -> s(f(0(), 1(), z)) }

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:
----------
  The following argument positions are usable:
    Uargs(s) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [f](x1, x2, x3) = [4] x3 + [1]
                                  
                [0] = [0]         
                                  
                [1] = [0]         
                                  
            [s](x1) = [1] x1 + [2]
  
  The following symbols are considered usable
  
    {f}
  
  The order satisfies the following ordering constraints:
  
     [f(x, y, s(z))] =  [4] z + [9]        
                     >  [4] z + [3]        
                     =  [s(f(0(), 1(), z))]
                                           
    [f(0(), 1(), x)] =  [4] x + [1]        
                     >= [4] x + [1]        
                     =  [f(s(x), x, 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(), 1(), x) -> f(s(x), x, x) }
Weak Trs: { f(x, y, s(z)) -> s(f(0(), 1(), z)) }
Obligation:
  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(), 1(), x) -> f(s(x), x, 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:
----------
  The following argument positions are usable:
    Uargs(s) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA) and not(IDA(1)).
  
    [f](x1, x2, x3) = [0 2] x1 + [2 0] x3 + [6]
                      [0 4]      [0 0]      [4]
                                               
                [0] = [0]                      
                      [1]                      
                                               
                [1] = [0]                      
                      [0]                      
                                               
            [s](x1) = [1 0] x1 + [2]           
                      [0 0]      [0]           
  
  The following symbols are considered usable
  
    {f}
  
  The order satisfies the following ordering constraints:
  
     [f(x, y, s(z))] =  [0 2] x + [2 0] z + [10]
                        [0 4]     [0 0]     [4] 
                     >= [2 0] z + [10]          
                        [0 0]     [0]           
                     =  [s(f(0(), 1(), z))]     
                                                
    [f(0(), 1(), x)] =  [2 0] x + [8]           
                        [0 0]     [8]           
                     >  [2 0] x + [6]           
                        [0 0]     [4]           
                     =  [f(s(x), x, 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(x, y, s(z)) -> s(f(0(), 1(), z))
  , f(0(), 1(), x) -> f(s(x), x, x) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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