YES

TRS:

  a(b(a(x))) -> b(a(x))

max/plus interpretations on N:

  a_A(x1) = max{0, x1}
  a#_A(x1) = max{0, x1}
  b_A(x1) = max{0, x1}
  b#_A(x1) = max{0, x1}

precedence:

  b > a