YES
O(n^3)
TRS:
 {
             sqr(0()) -> 0(),
             sqr(s()) -> s(),
             terms(N) -> cons(recip(sqr(N))),
             dbl(0()) -> 0(),
             dbl(s()) -> s(),
          add(0(), X) -> X,
          add(s(), Y) -> s(),
        first(0(), X) -> nil(),
  first(s(), cons(Y)) -> cons(Y)
 }
 DUP: We consider a non-duplicating system.
  Trs:
   {
               sqr(0()) -> 0(),
               sqr(s()) -> s(),
               terms(N) -> cons(recip(sqr(N))),
               dbl(0()) -> 0(),
               dbl(s()) -> s(),
            add(0(), X) -> X,
            add(s(), Y) -> s(),
          first(0(), X) -> nil(),
    first(s(), cons(Y)) -> cons(Y)
   }
  Matrix Interpretation:
   Interpretation class: triangular
           [X5]  [X2]    [1 0 0][X5]   [1 0 0][X2]   [0]
   [first]([X4], [X1]) = [0 0 0][X4] + [0 0 0][X1] + [0]
           [X3]  [X0]    [0 0 0][X3]   [0 0 0][X0]   [1]
   
           [0]
   [nil] = [0]
           [0]
   
         [X5]  [X2]    [1 1 1][X5]   [1 0 0][X2]   [1]
   [add]([X4], [X1]) = [0 0 0][X4] + [0 1 0][X1] + [1]
         [X3]  [X0]    [0 0 1][X3]   [0 0 1][X0]   [0]
   
         [X2]    [1 0 0][X2]   [1]
   [dbl]([X1]) = [0 0 0][X1] + [1]
         [X0]    [0 0 1][X0]   [1]
   
         [1]
   [s] = [1]
         [1]
   
         [1]
   [0] = [1]
         [1]
   
           [X2]    [1 1 0][X2]   [1]
   [terms]([X1]) = [0 0 0][X1] + [1]
           [X0]    [0 0 0][X0]   [0]
   
         [X2]    [1 1 0][X2]   [0]
   [sqr]([X1]) = [0 1 1][X1] + [0]
         [X0]    [0 0 0][X0]   [1]
   
           [X2]    [1 0 0][X2]   [0]
   [recip]([X1]) = [0 0 0][X1] + [0]
           [X0]    [0 0 0][X0]   [0]
   
          [X2]    [1 0 0][X2]   [0]
   [cons]([X1]) = [0 0 0][X1] + [0]
          [X0]    [0 0 0][X0]   [0]
   
   
   Qed