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)))

linear polynomial interpretations on N:

  minus_A(x1,x2) = x1
  minus#_A(x1,x2) = 0
  0_A = 1
  0#_A = 0
  s_A(x1) = x1 + 1
  s#_A(x1) = 1
  quot_A(x1,x2) = x1 + 1
  quot#_A(x1,x2) = x1 + x2

precedence:

  quot > minus = s > 0