YES

TRS:

  f(cons(nil(),y)) -> y
  f(cons(f(cons(nil(),y)),z)) -> copy(n(),y,z)
  copy(0(),y,z) -> f(z)
  copy(s(x),y,z) -> copy(x,y,cons(f(y),z))

max/plus interpretations on N:

  f_A(x1) = max{1, 2 + x1}
  f#_A(x1) = max{4, 0}
  cons_A(x1,x2) = max{2, 1, x2}
  cons#_A(x1,x2) = max{0, 1, 0}
  nil_A = 1
  nil#_A = 6
  copy_A(x1,x2,x3) = max{0, 4, 2, 2 + x3}
  copy#_A(x1,x2,x3) = max{0, 2 + x1, 0, 0}
  n_A = 1
  n#_A = 5
  0_A = 3
  0#_A = 0
  s_A(x1) = max{3, x1}
  s#_A(x1) = max{0, 0}

precedence:

  f = cons = nil = 0 = s > copy = n