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

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

Trs: { f(x, c(y)) -> f(x, s(f(y, 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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [f](x1, x2) = [1 0] x1 + [1 1] x2 + [0]
                  [0 0]      [0 0]      [0]
                                           
        [c](x1) = [1 0] x1 + [0]           
                  [1 1]      [1]           
                                           
        [s](x1) = [1 0] x1 + [0]           
                  [0 0]      [0]           
  
  This order satisfies the following ordering constraints:
  
    [f(x, c(y))] =  [1 0] x + [2 1] y + [1]
                    [0 0]     [0 0]     [0]
                 >  [1 0] x + [2 1] y + [0]
                    [0 0]     [0 0]     [0]
                 =  [f(x, s(f(y, y)))]     
                                           
    [f(s(x), y)] =  [1 0] x + [1 1] y + [0]
                    [0 0]     [0 0]     [0]
                 >= [1 0] x + [1 0] y + [0]
                    [0 0]     [0 0]     [0]
                 =  [f(x, s(c(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 Trs: { f(s(x), y) -> f(x, s(c(y))) }
Weak Trs: { f(x, c(y)) -> f(x, s(f(y, y))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

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

Trs: { f(s(x), y) -> f(x, s(c(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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [f](x1, x2) = [2 0] x1 + [1 2] x2 + [0]
                  [2 0]      [0 0]      [0]
                                           
        [c](x1) = [1 0] x1 + [0]           
                  [1 1]      [2]           
                                           
        [s](x1) = [1 0] x1 + [2]           
                  [0 0]      [0]           
  
  This order satisfies the following ordering constraints:
  
    [f(x, c(y))] = [2 0] x + [3 2] y + [4]
                   [2 0]     [0 0]     [0]
                 > [2 0] x + [3 2] y + [2]
                   [2 0]     [0 0]     [0]
                 = [f(x, s(f(y, y)))]     
                                          
    [f(s(x), y)] = [2 0] x + [1 2] y + [4]
                   [2 0]     [0 0]     [4]
                 > [2 0] x + [1 0] y + [2]
                   [2 0]     [0 0]     [0]
                 = [f(x, s(c(y)))]        
                                          

We return to the main proof.

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

Weak Trs:
  { f(x, c(y)) -> f(x, s(f(y, y)))
  , f(s(x), y) -> f(x, s(c(y))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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