YES

Problem:
 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)))

Proof:
 DP Processor:
  DPs:
   minus#(x,s(y)) -> minus#(x,y)
   minus#(x,s(y)) -> pred#(minus(x,y))
   quot#(s(x),s(y)) -> minus#(x,y)
   quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
  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)))
  Usable Rule Processor:
   DPs:
    minus#(x,s(y)) -> minus#(x,y)
    minus#(x,s(y)) -> pred#(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(x,s(y)) -> pred(minus(x,y))
    pred(s(x)) -> x
   Matrix Interpretation Processor: dim=1
    
    usable rules:
     minus(x,0()) -> x
     minus(x,s(y)) -> pred(minus(x,y))
     pred(s(x)) -> x
    interpretation:
     [quot#](x0, x1) = x0 + 4x1,
     
     [minus#](x0, x1) = 2x0 + 4x1 + 2,
     
     [pred#](x0) = 2x0 + 5,
     
     [minus](x0, x1) = x0,
     
     [0] = 0,
     
     [pred](x0) = x0,
     
     [s](x0) = 2x0 + 1
    orientation:
     minus#(x,s(y)) = 2x + 8y + 6 >= 2x + 4y + 2 = minus#(x,y)
     
     minus#(x,s(y)) = 2x + 8y + 6 >= 2x + 5 = pred#(minus(x,y))
     
     quot#(s(x),s(y)) = 2x + 8y + 5 >= 2x + 4y + 2 = minus#(x,y)
     
     quot#(s(x),s(y)) = 2x + 8y + 5 >= x + 8y + 4 = quot#(minus(x,y),s(y))
     
     minus(x,0()) = x >= x = x
     
     minus(x,s(y)) = x >= x = pred(minus(x,y))
     
     pred(s(x)) = 2x + 1 >= x = x
    problem:
     DPs:
      
     TRS:
      minus(x,0()) -> x
      minus(x,s(y)) -> pred(minus(x,y))
      pred(s(x)) -> x
    Qed