YES(O(1),O(n^2))

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

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^2))

We add following weak dependency pairs:

Strict DPs:
  { half^#(0()) -> c_1()
  , half^#(s(0())) -> c_2()
  , half^#(s(s(x))) -> c_3(half^#(x))
  , bits^#(0()) -> c_4()
  , bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict DPs:
  { half^#(0()) -> c_1()
  , half^#(s(0())) -> c_2()
  , half^#(s(s(x))) -> c_3(half^#(x))
  , bits^#(0()) -> c_4()
  , bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
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^2))

We replace rewrite rules by usable rules:

  Strict Usable Rules:
    { half(0()) -> 0()
    , half(s(0())) -> 0()
    , half(s(s(x))) -> s(half(x)) }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict DPs:
  { half^#(0()) -> c_1()
  , half^#(s(0())) -> c_2()
  , half^#(s(s(x))) -> c_3(half^#(x))
  , bits^#(0()) -> c_4()
  , bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Strict 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^2))

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

The following argument positions are usable:
  Uargs(s) = {1}, Uargs(c_3) = {1}, Uargs(bits^#) = {1},
  Uargs(c_5) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

    [half](x1) = [1] x1 + [2]
                             
           [0] = [1]         
                             
       [s](x1) = [1] x1 + [1]
                             
  [half^#](x1) = [2] x1 + [2]
                             
         [c_1] = [1]         
                             
         [c_2] = [2]         
                             
     [c_3](x1) = [1] x1 + [1]
                             
  [bits^#](x1) = [2] x1 + [1]
                             
         [c_4] = [2]         
                             
     [c_5](x1) = [1] x1 + [0]

This order satisfies following ordering constraints:

      [half(0())] = [3]         
                  > [1]         
                  = [0()]       
                                
   [half(s(0()))] = [4]         
                  > [1]         
                  = [0()]       
                                
  [half(s(s(x)))] = [1] x + [4] 
                  > [1] x + [3] 
                  = [s(half(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(n^2)).

Strict DPs: { bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Weak DPs:
  { half^#(0()) -> c_1()
  , half^#(s(0())) -> c_2()
  , half^#(s(s(x))) -> c_3(half^#(x))
  , bits^#(0()) -> c_4() }
Weak Trs:
  { half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ half^#(0()) -> c_1()
, half^#(s(0())) -> c_2()
, half^#(s(s(x))) -> c_3(half^#(x))
, bits^#(0()) -> c_4() }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^2)).

Strict DPs: { bits^#(s(x)) -> c_5(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(n^2))

The following argument positions are usable:
  Uargs(c_5) = {1}

TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).

    [half](x1) = [1 0] x1 + [0]
                 [1 0]      [0]
                               
           [0] = [0]           
                 [0]           
                               
       [s](x1) = [1 0] x1 + [1]
                 [1 0]      [2]
                               
  [bits^#](x1) = [1 2] x1 + [0]
                 [0 0]      [0]
                               
     [c_5](x1) = [1 0] x1 + [1]
                 [0 0]      [0]

This order satisfies following ordering constraints:

      [half(0())] =  [0]                      
                     [0]                      
                  >= [0]                      
                     [0]                      
                  =  [0()]                    
                                              
   [half(s(0()))] =  [1]                      
                     [1]                      
                  >  [0]                      
                     [0]                      
                  =  [0()]                    
                                              
  [half(s(s(x)))] =  [1 0] x + [2]            
                     [1 0]     [2]            
                  >  [1 0] x + [1]            
                     [1 0]     [2]            
                  =  [s(half(x))]             
                                              
   [bits^#(s(x))] =  [3 0] x + [5]            
                     [0 0]     [0]            
                  >  [3 0] x + [4]            
                     [0 0]     [0]            
                  =  [c_5(bits^#(half(s(x))))]
                                              

Hurray, we answered YES(O(1),O(n^2))