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)))
  TDG 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)))
   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#(x,s(y)) -> pred#(minus(x,y))
    quot#(s(x),s(y)) -> minus#(x,y) -> minus#(x,s(y)) -> minus#(x,y)
    minus#(x,s(y)) -> minus#(x,y) ->
    minus#(x,s(y)) -> pred#(minus(x,y))
    minus#(x,s(y)) -> minus#(x,y) -> minus#(x,s(y)) -> minus#(x,y)
   SCC Processor:
    #sccs: 2
    #rules: 2
    #arcs: 6/16
    DPs:
     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:
      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
     Arctic Interpretation Processor:
      dimension: 1
      usable rules:
       minus(x,0()) -> x
       minus(x,s(y)) -> pred(minus(x,y))
       pred(s(x)) -> x
      interpretation:
       [quot#](x0, x1) = x0,
       
       [minus](x0, x1) = x0 + -8,
       
       [0] = 4,
       
       [pred](x0) = x0 + -16,
       
       [s](x0) = 1x0 + 0
      orientation:
       quot#(s(x),s(y)) = 1x + 0 >= x + -8 = quot#(minus(x,y),s(y))
       
       minus(x,0()) = x + -8 >= x = x
       
       minus(x,s(y)) = x + -8 >= x + -8 = pred(minus(x,y))
       
       pred(s(x)) = 1x + 0 >= x = x
      problem:
       DPs:
        
       TRS:
        minus(x,0()) -> x
        minus(x,s(y)) -> pred(minus(x,y))
        pred(s(x)) -> x
      Qed
    
    DPs:
     minus#(x,s(y)) -> minus#(x,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)))
    Subterm Criterion Processor:
     simple projection:
      pi(minus#) = 1
     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