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)))
 plus(0(),y) -> y
 plus(s(x),y) -> s(plus(x,y))
 plus(minus(x,s(0())),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0())))

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))
   plus#(s(x),y) -> plus#(x,y)
   plus#(minus(x,s(0())),minus(y,s(s(z)))) -> plus#(minus(y,s(s(z))),minus(x,s(0())))
  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)))
   plus(0(),y) -> y
   plus(s(x),y) -> s(plus(x,y))
   plus(minus(x,s(0())),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(0())))
  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))
    plus#(s(x),y) -> plus#(x,y)
    plus#(minus(x,s(0())),minus(y,s(s(z)))) -> plus#(minus(y,s(s(z))),minus(x,s(0())))
   TRS:
    minus(x,0()) -> x
    minus(s(x),s(y)) -> minus(x,y)
   Matrix Interpretation Processor: dim=3
    
    usable rules:
     minus(x,0()) -> x
     minus(s(x),s(y)) -> minus(x,y)
    interpretation:
     [plus#](x0, x1) = [1 0 0]x0 + [0 1 0]x1,
     
     [quot#](x0, x1) = [0 0 1]x0,
     
     [minus#](x0, x1) = [1 0 0]x0,
     
               [1 0 1]     [1]
     [s](x0) = [1 1 0]x0 + [0]
               [1 0 1]     [1],
     
                       [1 1 1]     [0 0 0]  
     [minus](x0, x1) = [1 1 1]x0 + [0 1 0]x1
                       [1 0 1]     [0 0 0]  ,
     
           [0]
     [0] = [0]
           [1]
    orientation:
     minus#(s(x),s(y)) = [1 0 1]x + [1] >= [1 0 0]x = minus#(x,y)
     
     quot#(s(x),s(y)) = [1 0 1]x + [1] >= [1 0 0]x = minus#(x,y)
     
     quot#(s(x),s(y)) = [1 0 1]x + [1] >= [1 0 1]x = quot#(minus(x,y),s(y))
     
     plus#(s(x),y) = [1 0 1]x + [0 1 0]y + [1] >= [1 0 0]x + [0 1 0]y = plus#(x,y)
     
     plus#(minus(x,s(0())),minus(y,s(s(z)))) = [1 1 1]x + [1 1 1]y + [2 1 1]z + [1] >= [1 1 1]x + [1 1 1]y = plus#(minus(y,s(s(z))),minus(x,s(0())))
     
                    [1 1 1]          
     minus(x,0()) = [1 1 1]x >= x = x
                    [1 0 1]          
     
                        [3 1 2]    [0 0 0]    [2]    [1 1 1]    [0 0 0]              
     minus(s(x),s(y)) = [3 1 2]x + [1 1 0]y + [2] >= [1 1 1]x + [0 1 0]y = minus(x,y)
                        [2 0 2]    [0 0 0]    [2]    [1 0 1]    [0 0 0]              
    problem:
     DPs:
      
     TRS:
      minus(x,0()) -> x
      minus(s(x),s(y)) -> minus(x,y)
    Qed