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:
  { 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))

We add the following innermost 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^1)).

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^1))

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^1)).

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^1))

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 the following constructor-restricted matrix
interpretation.

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

The following symbols are considered usable

  {half, half^#, bits^#}

The order satisfies the following ordering constraints:

        [half(0())] = [3]                      
                      [1]                      
                    > [2]                      
                      [0]                      
                    = [0()]                    
                                               
     [half(s(0()))] = [5]                      
                      [1]                      
                    > [2]                      
                      [0]                      
                    = [0()]                    
                                               
    [half(s(s(x)))] = [1 1] x + [5]            
                      [0 0]     [1]            
                    > [1 0] x + [4]            
                      [0 0]     [0]            
                    = [s(half(x))]             
                                               
      [half^#(0())] = [0]                      
                      [0]                      
                    ? [1]                      
                      [1]                      
                    = [c_1()]                  
                                               
   [half^#(s(0()))] = [0]                      
                      [0]                      
                    ? [1]                      
                      [1]                      
                    = [c_2()]                  
                                               
  [half^#(s(s(x)))] = [0]                      
                      [0]                      
                    ? [2]                      
                      [2]                      
                    = [c_3(half^#(x))]         
                                               
      [bits^#(0())] = [4]                      
                      [0]                      
                    > [1]                      
                      [0]                      
                    = [c_4()]                  
                                               
     [bits^#(s(x))] = [2 2] x + [4]            
                      [0 0]     [0]            
                    ? [2 2] x + [7]            
                      [0 0]     [2]            
                    = [c_5(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 DPs:
  { half^#(0()) -> c_1()
  , half^#(s(0())) -> c_2()
  , half^#(s(s(x))) -> c_3(half^#(x))
  , bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Weak DPs: { 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(1),O(n^1))

We estimate the number of application of {1,2} by applications of
Pre({1,2}) = {3}. Here rules are labeled as follows:

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

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

Strict DPs:
  { half^#(s(s(x))) -> c_3(half^#(x))
  , bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Weak DPs:
  { half^#(0()) -> c_1()
  , half^#(s(0())) -> c_2()
  , 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(1),O(n^1))

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()
, bits^#(0()) -> c_4() }

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

Strict DPs:
  { half^#(s(s(x))) -> c_3(half^#(x))
  , 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(1),O(n^1))

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

DPs:
  { 1: half^#(s(s(x))) -> c_3(half^#(x)) }
Trs:
  { half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x)) }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_3) = {1}, Uargs(c_5) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
      [half](x1) = [1] x1 + [0]
                               
             [0] = [4]         
                               
         [s](x1) = [1] x1 + [4]
                               
      [bits](x1) = [7] x1 + [0]
                               
    [half^#](x1) = [1] x1 + [0]
                               
           [c_1] = [0]         
                               
           [c_2] = [0]         
                               
       [c_3](x1) = [1] x1 + [5]
                               
    [bits^#](x1) = [0]         
                               
           [c_4] = [0]         
                               
       [c_5](x1) = [2] x1 + [0]
  
  The following symbols are considered usable
  
    {half, half^#, bits^#}
  
  The order satisfies the following ordering constraints:
  
          [half(0())] =  [4]                      
                      >= [4]                      
                      =  [0()]                    
                                                  
       [half(s(0()))] =  [8]                      
                      >  [4]                      
                      =  [0()]                    
                                                  
      [half(s(s(x)))] =  [1] x + [8]              
                      >  [1] x + [4]              
                      =  [s(half(x))]             
                                                  
    [half^#(s(s(x)))] =  [1] x + [8]              
                      >  [1] x + [5]              
                      =  [c_3(half^#(x))]         
                                                  
       [bits^#(s(x))] =  [0]                      
                      >= [0]                      
                      =  [c_5(bits^#(half(s(x))))]
                                                  

The strictly oriented rules are moved into the weak component.

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

Strict DPs: { bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Weak DPs: { half^#(s(s(x))) -> c_3(half^#(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 following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ half^#(s(s(x))) -> c_3(half^#(x)) }

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

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(1),O(n^1))

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

DPs:
  { 1: bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Trs:
  { half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x)) }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_5) = {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] = [1]           
                   [0]           
                                 
         [s](x1) = [1 0] x1 + [1]
                   [1 0]      [2]
                                 
      [bits](x1) = [7 7] x1 + [0]
                   [0 0]      [0]
                                 
    [half^#](x1) = [7 7] x1 + [0]
                   [0 0]      [0]
                                 
           [c_1] = [0]           
                   [0]           
                                 
           [c_2] = [0]           
                   [0]           
                                 
       [c_3](x1) = [7 7] x1 + [0]
                   [0 0]      [0]
                                 
    [bits^#](x1) = [0 1] x1 + [0]
                   [0 0]      [4]
                                 
           [c_4] = [0]           
                   [0]           
                                 
       [c_5](x1) = [1 0] x1 + [0]
                   [0 0]      [3]
  
  The following symbols are considered usable
  
    {half, bits^#}
  
  The order satisfies the following ordering constraints:
  
        [half(0())] =  [1]                      
                       [1]                      
                    >= [1]                      
                       [0]                      
                    =  [0()]                    
                                                
     [half(s(0()))] =  [2]                      
                       [2]                      
                    >  [1]                      
                       [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))] =  [1 0] x + [2]            
                       [0 0]     [4]            
                    >  [1 0] x + [1]            
                       [0 0]     [3]            
                    =  [c_5(bits^#(half(s(x))))]
                                                

The strictly oriented rules are moved into the weak component.

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

Weak 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(1),O(1))

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

{ bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }

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)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

No rule is usable, rules are removed from the input problem.

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

Rules: Empty
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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