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:
  { f(a(), a()) -> f(a(), b())
  , f(a(), b()) -> f(s(a()), c())
  , f(s(X), c()) -> f(X, c())
  , f(c(), c()) -> f(a(), a()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We add following weak dependency pairs:

Strict DPs:
  { f^#(a(), a()) -> c_1(f^#(a(), b()))
  , f^#(a(), b()) -> c_2(f^#(s(a()), c()))
  , f^#(s(X), c()) -> c_3(f^#(X, c()))
  , f^#(c(), c()) -> c_4(f^#(a(), a())) }

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:
  { f^#(a(), a()) -> c_1(f^#(a(), b()))
  , f^#(a(), b()) -> c_2(f^#(s(a()), c()))
  , f^#(s(X), c()) -> c_3(f^#(X, c()))
  , f^#(c(), c()) -> c_4(f^#(a(), a())) }
Strict Trs:
  { f(a(), a()) -> f(a(), b())
  , f(a(), b()) -> f(s(a()), c())
  , f(s(X), c()) -> f(X, c())
  , f(c(), c()) -> f(a(), a()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

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(n^2)).

Strict DPs:
  { f^#(a(), a()) -> c_1(f^#(a(), b()))
  , f^#(a(), b()) -> c_2(f^#(s(a()), c()))
  , f^#(s(X), c()) -> c_3(f^#(X, c()))
  , f^#(c(), c()) -> c_4(f^#(a(), a())) }
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(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {1},
  Uargs(c_4) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

            [a] = [2]         
                              
            [b] = [0]         
                              
        [s](x1) = [1] x1 + [0]
                              
            [c] = [0]         
                              
  [f^#](x1, x2) = [2] x2 + [0]
                              
      [c_1](x1) = [1] x1 + [1]
                              
      [c_2](x1) = [1] x1 + [1]
                              
      [c_3](x1) = [1] x1 + [1]
                              
      [c_4](x1) = [1] x1 + [1]

This order satisfies following ordering constraints:


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:
  { f^#(a(), b()) -> c_2(f^#(s(a()), c()))
  , f^#(s(X), c()) -> c_3(f^#(X, c()))
  , f^#(c(), c()) -> c_4(f^#(a(), a())) }
Weak DPs: { f^#(a(), a()) -> c_1(f^#(a(), b())) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

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

  DPs:
    { 1: f^#(a(), b()) -> c_2(f^#(s(a()), c()))
    , 2: f^#(s(X), c()) -> c_3(f^#(X, c()))
    , 3: f^#(c(), c()) -> c_4(f^#(a(), a()))
    , 4: f^#(a(), a()) -> c_1(f^#(a(), b())) }

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

Strict DPs:
  { f^#(a(), b()) -> c_2(f^#(s(a()), c()))
  , f^#(s(X), c()) -> c_3(f^#(X, c())) }
Weak DPs:
  { f^#(a(), a()) -> c_1(f^#(a(), b()))
  , f^#(c(), c()) -> c_4(f^#(a(), a())) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^2))

The following argument positions are usable:
  Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {1},
  Uargs(c_4) = {1}

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

            [a] = [1]                      
                  [2]                      
                                           
            [b] = [0]                      
                  [2]                      
                                           
        [s](x1) = [1 0] x1 + [1]           
                  [0 0]      [0]           
                                           
            [c] = [3]                      
                  [0]                      
                                           
  [f^#](x1, x2) = [1 0] x1 + [0 1] x2 + [0]
                  [0 0]      [0 0]      [0]
                                           
      [c_1](x1) = [1 0] x1 + [0]           
                  [0 0]      [0]           
                                           
      [c_2](x1) = [1 0] x1 + [0]           
                  [0 0]      [0]           
                                           
      [c_3](x1) = [1 0] x1 + [0]           
                  [0 0]      [0]           
                                           
      [c_4](x1) = [1 0] x1 + [0]           
                  [0 0]      [0]           

This order satisfies following ordering constraints:

   [f^#(a(), a())] =  [3]                    
                      [0]                    
                   >= [3]                    
                      [0]                    
                   =  [c_1(f^#(a(), b()))]   
                                             
   [f^#(a(), b())] =  [3]                    
                      [0]                    
                   >  [2]                    
                      [0]                    
                   =  [c_2(f^#(s(a()), c()))]
                                             
  [f^#(s(X), c())] =  [1 0] X + [1]          
                      [0 0]     [0]          
                   >  [1 0] X + [0]          
                      [0 0]     [0]          
                   =  [c_3(f^#(X, c()))]     
                                             
   [f^#(c(), c())] =  [3]                    
                      [0]                    
                   >= [3]                    
                      [0]                    
                   =  [c_4(f^#(a(), a()))]   
                                             

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