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)))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [quot#](x0, x1) = x0 + 2x1,
    
    [minus#](x0, x1) = 1/2x0 + x1 + 1,
    
    [pred#](x0) = 1/2x0 + 1,
    
    [quot](x0, x1) = 2x0 + 7/2,
    
    [minus](x0, x1) = x0,
    
    [0] = 2,
    
    [pred](x0) = x0,
    
    [s](x0) = x0 + 1/2
   orientation:
    minus#(x,s(y)) = 1/2x + y + 3/2 >= 1/2x + y + 1 = minus#(x,y)
    
    minus#(x,s(y)) = 1/2x + y + 3/2 >= 1/2x + 1 = pred#(minus(x,y))
    
    quot#(s(x),s(y)) = x + 2y + 3/2 >= 1/2x + y + 1 = minus#(x,y)
    
    quot#(s(x),s(y)) = x + 2y + 3/2 >= x + 2y + 1 = quot#(minus(x,y),s(y))
    
    pred(s(x)) = x + 1/2 >= x = x
    
    minus(x,0()) = x >= x = x
    
    minus(x,s(y)) = x >= x = pred(minus(x,y))
    
    quot(0(),s(y)) = 15/2 >= 2 = 0()
    
    quot(s(x),s(y)) = 2x + 9/2 >= 2x + 4 = s(quot(minus(x,y),s(y)))
   problem:
    DPs:
     
    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)))
   Qed