We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).

Strict Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , bits(0()) -> 0()
  , bits(s(x)) -> s(bits(half(s(x)))) }
Obligation:
  innermost 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}, Uargs(bits) = {1}

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

  [half](x1) = [1] x1 + [0]
                           
         [0] = [0]         
                           
     [s](x1) = [1] x1 + [4]
                           
  [bits](x1) = [1] x1 + [0]

The order satisfies the following ordering constraints:

      [half(0())] =  [0]                  
                  >= [0]                  
                  =  [0()]                
                                          
   [half(s(0()))] =  [4]                  
                  >  [0]                  
                  =  [0()]                
                                          
  [half(s(s(x)))] =  [1] x + [8]          
                  >  [1] x + [4]          
                  =  [s(half(x))]         
                                          
      [bits(0())] =  [0]                  
                  >= [0]                  
                  =  [0()]                
                                          
     [bits(s(x))] =  [1] x + [4]          
                  ?  [1] x + [8]          
                  =  [s(bits(half(s(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:
  { half(0()) -> 0()
  , bits(0()) -> 0()
  , bits(s(x)) -> s(bits(half(s(x)))) }
Weak Trs:
  { half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x)) }
Obligation:
  innermost 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}, Uargs(bits) = {1}

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

  [half](x1) = [1] x1 + [4]
                           
         [0] = [4]         
                           
     [s](x1) = [1] x1 + [4]
                           
  [bits](x1) = [1] x1 + [0]

The order satisfies the following ordering constraints:

      [half(0())] =  [8]                  
                  >  [4]                  
                  =  [0()]                
                                          
   [half(s(0()))] =  [12]                 
                  >  [4]                  
                  =  [0()]                
                                          
  [half(s(s(x)))] =  [1] x + [12]         
                  >  [1] x + [8]          
                  =  [s(half(x))]         
                                          
      [bits(0())] =  [4]                  
                  >= [4]                  
                  =  [0()]                
                                          
     [bits(s(x))] =  [1] x + [4]          
                  ?  [1] x + [12]         
                  =  [s(bits(half(s(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:
  { bits(0()) -> 0()
  , bits(s(x)) -> s(bits(half(s(x)))) }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x)) }
Obligation:
  innermost 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}, Uargs(bits) = {1}

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

  [half](x1) = [7]         
                           
         [0] = [0]         
                           
     [s](x1) = [1] x1 + [0]
                           
  [bits](x1) = [1] x1 + [1]

The order satisfies the following ordering constraints:

      [half(0())] =  [7]                  
                  >  [0]                  
                  =  [0()]                
                                          
   [half(s(0()))] =  [7]                  
                  >  [0]                  
                  =  [0()]                
                                          
  [half(s(s(x)))] =  [7]                  
                  >= [7]                  
                  =  [s(half(x))]         
                                          
      [bits(0())] =  [1]                  
                  >  [0]                  
                  =  [0()]                
                                          
     [bits(s(x))] =  [1] x + [1]          
                  ?  [8]                  
                  =  [s(bits(half(s(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: { bits(s(x)) -> s(bits(half(s(x)))) }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , bits(0()) -> 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: { bits(s(x)) -> s(bits(half(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:
----------
  The following argument positions are usable:
    Uargs(s) = {1}, Uargs(bits) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA) and not(IDA(1)).
  
    [half](x1) = [1 0] x1 + [0]
                 [1 0]      [0]
                               
           [0] = [0]           
                 [0]           
                               
       [s](x1) = [1 0] x1 + [2]
                 [1 0]      [4]
                               
    [bits](x1) = [1 3] x1 + [1]
                 [3 1]      [3]
  
  The order satisfies the following ordering constraints:
  
        [half(0())] =  [0]                  
                       [0]                  
                    >= [0]                  
                       [0]                  
                    =  [0()]                
                                            
     [half(s(0()))] =  [2]                  
                       [2]                  
                    >  [0]                  
                       [0]                  
                    =  [0()]                
                                            
    [half(s(s(x)))] =  [1 0] x + [4]        
                       [1 0]     [4]        
                    >  [1 0] x + [2]        
                       [1 0]     [4]        
                    =  [s(half(x))]         
                                            
        [bits(0())] =  [1]                  
                       [3]                  
                    >  [0]                  
                       [0]                  
                    =  [0()]                
                                            
       [bits(s(x))] =  [4 0] x + [15]       
                       [4 0]     [13]       
                    >  [4 0] x + [11]       
                       [4 0]     [13]       
                    =  [s(bits(half(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:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , bits(0()) -> 0()
  , bits(s(x)) -> s(bits(half(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))