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(s(x))) -> s(half(x))
  , log(s(0())) -> 0()
  , log(s(s(x))) -> s(log(s(half(x)))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add following weak dependency pairs:

Strict DPs:
  { half^#(0()) -> c_1()
  , half^#(s(s(x))) -> c_2(half^#(x))
  , log^#(s(0())) -> c_3()
  , log^#(s(s(x))) -> c_4(log^#(s(half(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(s(x))) -> c_2(half^#(x))
  , log^#(s(0())) -> c_3()
  , log^#(s(s(x))) -> c_4(log^#(s(half(x)))) }
Strict Trs:
  { half(0()) -> 0()
  , half(s(s(x))) -> s(half(x))
  , log(s(0())) -> 0()
  , log(s(s(x))) -> s(log(s(half(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(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(s(x))) -> c_2(half^#(x))
  , log^#(s(0())) -> c_3()
  , log^#(s(s(x))) -> c_4(log^#(s(half(x)))) }
Strict Trs:
  { half(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_2) = {1}, Uargs(log^#) = {1},
  Uargs(c_4) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

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

This order satisfies following ordering constraints:

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

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

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

Strict DPs: { log^#(s(s(x))) -> c_4(log^#(s(half(x)))) }
Weak Trs:
  { half(0()) -> 0()
  , half(s(s(x))) -> s(half(x)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The following argument positions are usable:
  Uargs(c_4) = {1}

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

   [half](x1) = [1] x1 + [0]
                            
          [0] = [3]         
                            
      [s](x1) = [1] x1 + [1]
                            
  [log^#](x1) = [2] x1 + [1]
                            
    [c_4](x1) = [1] x1 + [1]

This order satisfies following ordering constraints:

       [half(0())] =  [3]                     
                   >= [3]                     
                   =  [0()]                   
                                              
   [half(s(s(x)))] =  [1] x + [2]             
                   >  [1] x + [1]             
                   =  [s(half(x))]            
                                              
  [log^#(s(s(x)))] =  [2] x + [5]             
                   >  [2] x + [4]             
                   =  [c_4(log^#(s(half(x))))]
                                              

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