YES

TRS:

  rev(a()) -> a()
  rev(b()) -> b()
  rev(++(x,y)) -> ++(rev(y),rev(x))
  rev(++(x,x)) -> rev(x)

max/plus interpretations on N:

  rev_A(x1) = max{4, 2 + x1}
  rev#_A(x1) = max{4, 2 + x1}
  a_A = 0
  a#_A = 0
  b_A = 0
  b#_A = 0
  ++_A(x1,x2) = max{8, 3 + x1, 3 + x2}
  ++#_A(x1,x2) = max{8, 3 + x1, 3 + x2}

precedence:

  a = b > rev = ++