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

Strict Trs:
  { f(X, X) -> c(X)
  , f(X, c(X)) -> f(s(X), X)
  , f(s(X), X) -> f(X, a(X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

We add the following weak dependency pairs:

Strict DPs:
  { f^#(X, X) -> c_1()
  , f^#(X, c(X)) -> c_2(f^#(s(X), X))
  , f^#(s(X), X) -> c_3(f^#(X, a(X))) }

and mark the set of starting terms.

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

Strict DPs:
  { f^#(X, X) -> c_1()
  , f^#(X, c(X)) -> c_2(f^#(s(X), X))
  , f^#(s(X), X) -> c_3(f^#(X, a(X))) }
Strict Trs:
  { f(X, X) -> c(X)
  , f(X, c(X)) -> f(s(X), X)
  , f(s(X), X) -> f(X, a(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)).

Strict DPs:
  { f^#(X, X) -> c_1()
  , f^#(X, c(X)) -> c_2(f^#(s(X), X))
  , f^#(s(X), X) -> c_3(f^#(X, a(X))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

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

The following argument positions are usable:
  Uargs(c_2) = {1}

TcT has computed the following constructor-restricted matrix
interpretation.

        [s](x1) = [0]           
                  [0]           
                                
        [a](x1) = [0]           
                  [0]           
                                
        [c](x1) = [0]           
                  [0]           
                                
  [f^#](x1, x2) = [1]           
                  [0]           
                                
          [c_1] = [0]           
                  [0]           
                                
      [c_2](x1) = [1 0] x1 + [0]
                  [0 1]      [0]
                                
      [c_3](x1) = [0]           
                  [0]           

The order satisfies the following ordering constraints:

     [f^#(X, X)] =  [1]                
                    [0]                
                 >  [0]                
                    [0]                
                 =  [c_1()]            
                                       
  [f^#(X, c(X))] =  [1]                
                    [0]                
                 >= [1]                
                    [0]                
                 =  [c_2(f^#(s(X), X))]
                                       
  [f^#(s(X), X)] =  [1]                
                    [0]                
                 >  [0]                
                    [0]                
                 =  [c_3(f^#(X, a(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(1)).

Strict DPs: { f^#(X, c(X)) -> c_2(f^#(s(X), X)) }
Weak DPs:
  { f^#(X, X) -> c_1()
  , f^#(s(X), X) -> c_3(f^#(X, a(X))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

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

  DPs:
    { 1: f^#(X, c(X)) -> c_2(f^#(s(X), X))
    , 2: f^#(X, X) -> c_1()
    , 3: f^#(s(X), X) -> c_3(f^#(X, a(X))) }

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

Weak DPs:
  { f^#(X, X) -> c_1()
  , f^#(X, c(X)) -> c_2(f^#(s(X), X))
  , f^#(s(X), X) -> c_3(f^#(X, a(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.

{ f^#(X, X) -> c_1()
, f^#(X, c(X)) -> c_2(f^#(s(X), X))
, f^#(s(X), X) -> c_3(f^#(X, a(X))) }

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(1))