YES

TRS:

  terms(N) -> cons(recip(sqr(N)))
  sqr(0()) -> 0()
  sqr(s()) -> s()
  dbl(0()) -> 0()
  dbl(s()) -> s()
  add(0(),X) -> X
  add(s(),Y) -> s()
  first(0(),X) -> nil()
  first(s(),cons(Y)) -> cons(Y)

max/plus interpretations on N:

  terms_A(x1) = max{5, 4 + x1}
  terms#_A(x1) = max{5, 4 + x1}
  cons_A(x1) = max{5, x1}
  cons#_A(x1) = max{5, x1}
  recip_A(x1) = max{5, 2 + x1}
  recip#_A(x1) = max{5, 2 + x1}
  sqr_A(x1) = max{3, 1 + x1}
  sqr#_A(x1) = max{3, 1 + x1}
  0_A = 1
  0#_A = 1
  s_A = 2
  s#_A = 2
  dbl_A(x1) = max{3, 1 + x1}
  dbl#_A(x1) = max{3, 1 + x1}
  add_A(x1,x2) = max{3, 1 + x1, 3 + x2}
  add#_A(x1,x2) = max{3, 1 + x1, 3 + x2}
  first_A(x1,x2) = max{6, 4 + x1, 6 + x2}
  first#_A(x1,x2) = max{6, 4 + x1, 6 + x2}
  nil_A = 0
  nil#_A = 0

precedence:

  0 > s = nil > terms = dbl = add = first > cons = recip = sqr