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(x, c(y)) -> f(x, s(f(y, y)))
  , f(s(x), s(y)) -> f(x, s(c(s(y)))) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^2))

We add the following weak dependency pairs:

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

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^#(x, c(y)) -> c_1(f^#(x, s(f(y, y))))
  , f^#(s(x), s(y)) -> c_2(f^#(x, s(c(s(y))))) }
Strict Trs:
  { f(x, c(y)) -> f(x, s(f(y, y)))
  , f(s(x), s(y)) -> f(x, s(c(s(y)))) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^2))

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

DPs: { f^#(x, c(y)) -> c_1(f^#(x, s(f(y, y)))) }
Trs:
  { f(x, c(y)) -> f(x, s(f(y, y)))
  , f(s(x), s(y)) -> f(x, s(c(s(y)))) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(f) = {1, 2}, Uargs(c) = {1}, Uargs(s) = {1},
    Uargs(f^#) = {2}, Uargs(c_1) = {1}, Uargs(c_2) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
      [f](x1, x2) = [2 0] x1 + [1 2] x2 + [1]
                    [0 0]      [0 0]      [0]
                                             
          [c](x1) = [1 0] x1 + [0]           
                    [1 1]      [4]           
                                             
          [s](x1) = [1 0] x1 + [1]           
                    [0 0]      [0]           
                                             
    [f^#](x1, x2) = [1 0] x1 + [1 2] x2 + [0]
                    [0 0]      [0 0]      [0]
                                             
        [c_1](x1) = [1 0] x1 + [1]           
                    [0 0]      [0]           
                                             
        [c_2](x1) = [1 0] x1 + [0]           
                    [0 0]      [0]           
  
  The following symbols are considered usable
  
    {f, f^#}
  
  The order satisfies the following ordering constraints:
  
         [f(x, c(y))] =  [2 0] x + [3 2] y + [9]  
                         [0 0]     [0 0]     [0]  
                      >  [2 0] x + [3 2] y + [3]  
                         [0 0]     [0 0]     [0]  
                      =  [f(x, s(f(y, y)))]       
                                                  
      [f(s(x), s(y))] =  [2 0] x + [1 0] y + [4]  
                         [0 0]     [0 0]     [0]  
                      >  [2 0] x + [1 0] y + [3]  
                         [0 0]     [0 0]     [0]  
                      =  [f(x, s(c(s(y))))]       
                                                  
       [f^#(x, c(y))] =  [1 0] x + [3 2] y + [8]  
                         [0 0]     [0 0]     [0]  
                      >  [1 0] x + [3 2] y + [3]  
                         [0 0]     [0 0]     [0]  
                      =  [c_1(f^#(x, s(f(y, y))))]
                                                  
    [f^#(s(x), s(y))] =  [1 0] x + [1 0] y + [2]  
                         [0 0]     [0 0]     [0]  
                      >= [1 0] x + [1 0] y + [2]  
                         [0 0]     [0 0]     [0]  
                      =  [c_2(f^#(x, s(c(s(y)))))]
                                                  

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict DPs: { f^#(s(x), s(y)) -> c_2(f^#(x, s(c(s(y))))) }
Weak DPs: { f^#(x, c(y)) -> c_1(f^#(x, s(f(y, y)))) }
Weak Trs:
  { f(x, c(y)) -> f(x, s(f(y, y)))
  , f(s(x), s(y)) -> f(x, s(c(s(y)))) }
Obligation:
  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(f) = {1, 2}, Uargs(c) = {1}, Uargs(s) = {1},
  Uargs(f^#) = {2}, Uargs(c_1) = {1}, Uargs(c_2) = {1}

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

    [f](x1, x2) = [1 0] x1 + [1 1] x2 + [1]
                  [0 0]      [0 0]      [0]
                                           
        [c](x1) = [1 0] x1 + [0]           
                  [1 1]      [3]           
                                           
        [s](x1) = [1 0] x1 + [1]           
                  [0 0]      [1]           
                                           
  [f^#](x1, x2) = [5 0] x1 + [4 4] x2 + [0]
                  [0 0]      [0 0]      [0]
                                           
      [c_1](x1) = [1 0] x1 + [0]           
                  [0 1]      [0]           
                                           
      [c_2](x1) = [1 0] x1 + [0]           
                  [0 1]      [0]           

The following symbols are considered usable

  {f, f^#}

The order satisfies the following ordering constraints:

       [f(x, c(y))] =  [1 0] x + [2 1] y + [4]  
                       [0 0]     [0 0]     [0]  
                    >= [1 0] x + [2 1] y + [4]  
                       [0 0]     [0 0]     [0]  
                    =  [f(x, s(f(y, y)))]       
                                                
    [f(s(x), s(y))] =  [1 0] x + [1 0] y + [4]  
                       [0 0]     [0 0]     [0]  
                    >= [1 0] x + [1 0] y + [4]  
                       [0 0]     [0 0]     [0]  
                    =  [f(x, s(c(s(y))))]       
                                                
     [f^#(x, c(y))] =  [5 0] x + [8 4] y + [12] 
                       [0 0]     [0 0]     [0]  
                    >= [5 0] x + [8 4] y + [12] 
                       [0 0]     [0 0]     [0]  
                    =  [c_1(f^#(x, s(f(y, y))))]
                                                
  [f^#(s(x), s(y))] =  [5 0] x + [4 0] y + [13] 
                       [0 0]     [0 0]     [0]  
                    >  [5 0] x + [4 0] y + [12] 
                       [0 0]     [0 0]     [0]  
                    =  [c_2(f^#(x, s(c(s(y)))))]
                                                

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)).

Weak DPs:
  { f^#(x, c(y)) -> c_1(f^#(x, s(f(y, y))))
  , f^#(s(x), s(y)) -> c_2(f^#(x, s(c(s(y))))) }
Weak Trs:
  { f(x, c(y)) -> f(x, s(f(y, y)))
  , f(s(x), s(y)) -> f(x, s(c(s(y)))) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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