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)))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [quot#](x0, x1) = 4x0 + 4x1,
    
    [minus#](x0, x1) = x1,
    
    [quot](x0, x1) = 3x0 + 6,
    
    [s](x0) = 2x0 + 3,
    
    [minus](x0, x1) = x0,
    
    [0] = 2
   orientation:
    minus#(s(x),s(y)) = 2y + 3 >= y = minus#(x,y)
    
    quot#(s(x),s(y)) = 8x + 8y + 24 >= y = minus#(x,y)
    
    quot#(s(x),s(y)) = 8x + 8y + 24 >= 4x + 8y + 12 = quot#(minus(x,y),s(y))
    
    minus(x,0()) = x >= x = x
    
    minus(s(x),s(y)) = 2x + 3 >= x = minus(x,y)
    
    quot(0(),s(y)) = 12 >= 2 = 0()
    
    quot(s(x),s(y)) = 6x + 15 >= 6x + 15 = s(quot(minus(x,y),s(y)))
   problem:
    DPs:
     
    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)))
   Qed