YES

Problem:
 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)))

Proof:
 DP Processor:
  DPs:
   minus#(s(x),s(y)) -> minus#(x,y)
   quot#(s(x),s(y)) -> minus#(x,y)
   quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
  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)))
  Usable Rule Processor:
   DPs:
    minus#(s(x),s(y)) -> minus#(x,y)
    quot#(s(x),s(y)) -> minus#(x,y)
    quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
   TRS:
    minus(x,0()) -> x
    minus(s(x),s(y)) -> minus(x,y)
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     minus(x,0()) -> x
     minus(s(x),s(y)) -> minus(x,y)
    interpretation:
     [quot#](x0, x1) = 4x0,
     
     [minus#](x0, x1) = x0,
     
     [s](x0) = 2x0 + 0,
     
     [minus](x0, x1) = 1x0,
     
     [0] = 1
    orientation:
     minus#(s(x),s(y)) = 2x + 0 >= x = minus#(x,y)
     
     quot#(s(x),s(y)) = 6x + 4 >= x = minus#(x,y)
     
     quot#(s(x),s(y)) = 6x + 4 >= 5x = quot#(minus(x,y),s(y))
     
     minus(x,0()) = 1x >= x = x
     
     minus(s(x),s(y)) = 3x + 1 >= 1x = minus(x,y)
    problem:
     DPs:
      
     TRS:
      minus(x,0()) -> x
      minus(s(x),s(y)) -> minus(x,y)
    Qed