YES

TRS:

  minus(x,0()) -> x
  minus(s(x),s(y)) -> 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:

  minus_A(x1,x2) = max{4, x1, -6}
  minus#_A(x1,x2) = max{0, 2, 1}
  0_A = 10
  0#_A = 7
  s_A(x1) = max{11, 7 + x1}
  s#_A(x1) = max{6, 2 + x1}
  quot_A(x1,x2) = max{7, 3 + x1, 4}
  quot#_A(x1,x2) = max{3, -2 + x1, 4}

precedence:

  0 > quot > minus = s