YES

TRS:

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

max/plus interpretations on N:

  terms_A(x1) = max{6, -3}
  terms#_A(x1) = max{11, 5}
  cons_A(x1) = max{6, -1 + x1}
  cons#_A(x1) = max{6, 11}
  recip_A(x1) = max{1, -1}
  recip#_A(x1) = max{10, 2 + x1}
  sqr_A(x1) = max{4, -1}
  sqr#_A(x1) = max{10, 4}
  0_A = 4
  0#_A = 9
  s_A(x1) = max{4, 2}
  s#_A(x1) = max{0, 2}
  add_A(x1,x2) = max{2, 4, 1 + x2}
  add#_A(x1,x2) = max{9, 5, -1}
  dbl_A(x1) = max{2, 1 + x1}
  dbl#_A(x1) = max{1, 3}
  first_A(x1,x2) = max{0, 1, x2}
  first#_A(x1,x2) = max{7, 0, 6 + x2}
  nil_A = 1
  nil#_A = 8
  half_A(x1) = max{3, x1}
  half#_A(x1) = max{7, 3}

precedence:

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