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, y) -> x
  , g(x, y) -> y
  , 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))

The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(s) = {1}

TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).

      [g](x1, x2) = [1] x1 + [1] x2 + [7]
                                         
  [f](x1, x2, x3) = [1] x3 + [7]         
                                         
              [0] = [7]                  
                                         
              [1] = [7]                  
                                         
          [s](x1) = [1] x1 + [7]         

The following symbols are considered usable

  {g, f}

The order satisfies the following ordering constraints:

         [g(x, y)] =  [1] x + [1] y + [7]
                   >  [1] x + [0]        
                   =  [x]                
                                         
         [g(x, y)] =  [1] x + [1] y + [7]
                   >  [1] y + [0]        
                   =  [y]                
                                         
   [f(x, y, s(z))] =  [1] z + [14]       
                   >= [1] z + [14]       
                   =  [s(f(0(), 1(), z))]
                                         
  [f(0(), 1(), x)] =  [1] x + [7]        
                   >= [1] x + [7]        
                   =  [f(s(x), x, x)]    
                                         

Further, it can be verified that all rules not oriented are covered by the weightgap condition.

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) }
Weak Trs:
  { g(x, y) -> x
  , g(x, y) -> y }
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).
  
        [g](x1, x2) = [7] x1 + [7] x2 + [7]
                                           
    [f](x1, x2, x3) = [4] x3 + [5]         
                                           
                [0] = [0]                  
                                           
                [1] = [0]                  
                                           
            [s](x1) = [1] x1 + [2]         
  
  The following symbols are considered usable
  
    {g, f}
  
  The order satisfies the following ordering constraints:
  
           [g(x, y)] =  [7] x + [7] y + [7]
                     >  [1] x + [0]        
                     =  [x]                
                                           
           [g(x, y)] =  [7] x + [7] y + [7]
                     >  [1] y + [0]        
                     =  [y]                
                                           
     [f(x, y, s(z))] =  [4] z + [13]       
                     >  [4] z + [7]        
                     =  [s(f(0(), 1(), z))]
                                           
    [f(0(), 1(), x)] =  [4] x + [5]        
                     >= [4] x + [5]        
                     =  [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:
  { g(x, y) -> x
  , g(x, y) -> y
  , 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)).
  
        [g](x1, x2) = [7 7] x1 + [7 7] x2 + [7]
                      [0 7]      [0 7]      [0]
                                               
    [f](x1, x2, x3) = [0 2] x1 + [4 2] x3 + [0]
                      [0 0]      [0 1]      [0]
                                               
                [0] = [0]                      
                      [4]                      
                                               
                [1] = [0]                      
                      [0]                      
                                               
            [s](x1) = [1 1] x1 + [2]           
                      [0 0]      [1]           
  
  The following symbols are considered usable
  
    {g, f}
  
  The order satisfies the following ordering constraints:
  
           [g(x, y)] =  [7 7] x + [7 7] y + [7] 
                        [0 7]     [0 7]     [0] 
                     >  [1 0] x + [0]           
                        [0 1]     [0]           
                     =  [x]                     
                                                
           [g(x, y)] =  [7 7] x + [7 7] y + [7] 
                        [0 7]     [0 7]     [0] 
                     >  [1 0] y + [0]           
                        [0 1]     [0]           
                     =  [y]                     
                                                
     [f(x, y, s(z))] =  [0 2] x + [4 4] z + [10]
                        [0 0]     [0 0]     [1] 
                     >= [4 3] z + [10]          
                        [0 0]     [1]           
                     =  [s(f(0(), 1(), z))]     
                                                
    [f(0(), 1(), x)] =  [4 2] x + [8]           
                        [0 1]     [0]           
                     >  [4 2] x + [2]           
                        [0 1]     [0]           
                     =  [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:
  { g(x, y) -> x
  , g(x, y) -> y
  , 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))