YES

TRS:

  pred(s(x)) -> x
  minus(x,0()) -> x
  minus(x,s(y)) -> pred(minus(x,y))
  quot(0(),s(y)) -> 0()
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))

max/plus interpretations on N:

  pred_A(x1) = max{1, x1}
  pred#_A(x1) = max{2, 2}
  s_A(x1) = max{8, 4 + x1}
  s#_A(x1) = max{0, 2}
  minus_A(x1,x2) = max{1, x1, -2}
  minus#_A(x1,x2) = max{1, 2, 2}
  0_A = 6
  0#_A = 1
  quot_A(x1,x2) = max{5, 1 + x1, -1}
  quot#_A(x1,x2) = max{3, 1 + x1, 4}

precedence:

  quot > minus = 0 > pred = s