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)))
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [quot#](x0, x1) = 1x0,
      
      [quot](x0, x1) = x0,
      
      [minus](x0, x1) = x0,
      
      [0] = 3,
      
      [pred](x0) = x0,
      
      [s](x0) = 1x0
     orientation:
      quot#(s(x),s(y)) = 2x >= 1x = quot#(minus(x,y),s(y))
      
      pred(s(x)) = 1x >= x = x
      
      minus(x,0()) = x >= x = x
      
      minus(x,s(y)) = x >= x = pred(minus(x,y))
      
      quot(0(),s(y)) = 3 >= 3 = 0()
      
      quot(s(x),s(y)) = 1x >= 1x = 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
    
    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