YES

TRS:

  rev(ls) -> r1(ls,empty())
  r1(empty(),a) -> a
  r1(cons(x,k),a) -> r1(k,cons(x,a))

max/plus interpretations on N:

  rev_A(x1) = max{4, 2 + x1}
  rev#_A(x1) = max{4, 2 + x1}
  r1_A(x1,x2) = max{4, 1 + x1, 1 + x2}
  r1#_A(x1,x2) = max{4, 1 + x1, 1 + x2}
  empty_A = 3
  empty#_A = 3
  cons_A(x1,x2) = max{0, x1, x2}
  cons#_A(x1,x2) = max{0, x1, x2}

precedence:

  rev > r1 = empty > cons