YES(?,O(n^2))

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

Strict Trs:
  { f(0()) -> s(0())
  , f(s(0())) -> s(0())
  , f(s(s(x))) -> f(f(s(x))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

We add following dependency tuples:

Strict DPs:
  { f^#(0()) -> c_1()
  , f^#(s(0())) -> c_2()
  , f^#(s(s(x))) -> c_3(f^#(f(s(x))), f^#(s(x))) }

and mark the set of starting terms.

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

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

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

  DPs:
    { 1: f^#(0()) -> c_1()
    , 2: f^#(s(0())) -> c_2()
    , 3: f^#(s(s(x))) -> c_3(f^#(f(s(x))), f^#(s(x))) }

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

Strict DPs: { f^#(s(s(x))) -> c_3(f^#(f(s(x))), f^#(s(x))) }
Weak DPs:
  { f^#(0()) -> c_1()
  , f^#(s(0())) -> c_2() }
Weak Trs:
  { f(0()) -> s(0())
  , f(s(0())) -> s(0())
  , f(s(s(x))) -> f(f(s(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.

{ f^#(0()) -> c_1()
, f^#(s(0())) -> c_2() }

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

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

The following argument positions are usable:
  Uargs(c_3) = {1, 2}

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

        [f](x1) = [2]                      
                  [0]                      
                                           
            [0] = [0]                      
                  [0]                      
                                           
        [s](x1) = [1 0] x1 + [1]           
                  [1 0]      [0]           
                                           
      [f^#](x1) = [0 1] x1 + [0]           
                  [0 0]      [0]           
                                           
  [c_3](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                  [0 0]      [0 0]      [0]

This order satisfies following ordering constraints:

        [f(0())] =  [2]                           
                    [0]                           
                 >  [1]                           
                    [0]                           
                 =  [s(0())]                      
                                                  
     [f(s(0()))] =  [2]                           
                    [0]                           
                 >  [1]                           
                    [0]                           
                 =  [s(0())]                      
                                                  
    [f(s(s(x)))] =  [2]                           
                    [0]                           
                 >= [2]                           
                    [0]                           
                 =  [f(f(s(x)))]                  
                                                  
  [f^#(s(s(x)))] =  [1 0] x + [1]                 
                    [0 0]     [0]                 
                 >  [1 0] x + [0]                 
                    [0 0]     [0]                 
                 =  [c_3(f^#(f(s(x))), f^#(s(x)))]
                                                  

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