YES

TRS:

  g(b()) -> f(b())
  f(a()) -> g(a())
  b() -> a()

max/plus interpretations on N:

  g_A(x1) = max{1, 0}
  g#_A(x1) = max{0, -1 + x1}
  b_A = 3
  b#_A = 3
  f_A(x1) = max{1, -1}
  f#_A(x1) = max{1, -1}
  a_A = 1
  a#_A = 0

precedence:

  g > b = f > a