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)))
  TDG 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)))
   graph:
    quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) ->
    quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
    quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) ->
    quot#(s(x),s(y)) -> minus#(x,y)
    quot#(s(x),s(y)) -> minus#(x,y) ->
    minus#(s(x),s(y)) -> minus#(x,y)
    minus#(s(x),s(y)) -> minus#(x,y) -> minus#(s(x),s(y)) -> minus#(x,y)
   SCC Processor:
    #sccs: 2
    #rules: 2
    #arcs: 4/9
    DPs:
     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)))
    KBO Processor:
     argument filtering:
      pi(0) = []
      pi(minus) = 0
      pi(s) = [0]
      pi(quot) = [0]
      pi(quot#) = 0
     weight function:
      w0 = 1
      w(quot#) = w(quot) = w(s) = w(minus) = w(0) = 1
     precedence:
      minus > quot# ~ quot ~ 0 > s
     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
    
    DPs:
     minus#(s(x),s(y)) -> minus#(x,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)))
    KBO Processor:
     argument filtering:
      pi(0) = []
      pi(minus) = 0
      pi(s) = [0]
      pi(quot) = 0
      pi(minus#) = 0
     weight function:
      w0 = 1
      w(minus#) = w(quot) = w(s) = w(minus) = w(0) = 1
     precedence:
      minus > minus# ~ quot ~ s ~ 0
     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