YES

TRS:

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

max/plus interpretations on N:

  a_A(x1) = max{0, 7 + x1}
  a#_A(x1) = max{0, 7 + x1}
  b_A(x1) = max{2, 4 + x1}
  b#_A(x1) = max{2, 4 + x1}

precedence:

  b > a