YES

Problem:
 minus(X,0()) -> X
 minus(s(X),s(Y)) -> p(minus(X,Y))
 p(s(X)) -> X
 div(0(),s(Y)) -> 0()
 div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))

Proof:
 DP Processor:
  DPs:
   minus#(s(X),s(Y)) -> minus#(X,Y)
   minus#(s(X),s(Y)) -> p#(minus(X,Y))
   div#(s(X),s(Y)) -> minus#(X,Y)
   div#(s(X),s(Y)) -> div#(minus(X,Y),s(Y))
  TRS:
   minus(X,0()) -> X
   minus(s(X),s(Y)) -> p(minus(X,Y))
   p(s(X)) -> X
   div(0(),s(Y)) -> 0()
   div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))
  TDG Processor:
   DPs:
    minus#(s(X),s(Y)) -> minus#(X,Y)
    minus#(s(X),s(Y)) -> p#(minus(X,Y))
    div#(s(X),s(Y)) -> minus#(X,Y)
    div#(s(X),s(Y)) -> div#(minus(X,Y),s(Y))
   TRS:
    minus(X,0()) -> X
    minus(s(X),s(Y)) -> p(minus(X,Y))
    p(s(X)) -> X
    div(0(),s(Y)) -> 0()
    div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))
   graph:
    div#(s(X),s(Y)) -> div#(minus(X,Y),s(Y)) ->
    div#(s(X),s(Y)) -> div#(minus(X,Y),s(Y))
    div#(s(X),s(Y)) -> div#(minus(X,Y),s(Y)) ->
    div#(s(X),s(Y)) -> minus#(X,Y)
    div#(s(X),s(Y)) -> minus#(X,Y) ->
    minus#(s(X),s(Y)) -> p#(minus(X,Y))
    div#(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)) -> p#(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: 6/16
    DPs:
     div#(s(X),s(Y)) -> div#(minus(X,Y),s(Y))
    TRS:
     minus(X,0()) -> X
     minus(s(X),s(Y)) -> p(minus(X,Y))
     p(s(X)) -> X
     div(0(),s(Y)) -> 0()
     div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [div#](x0, x1) = 4x0,
      
      [div](x0, x1) = x0,
      
      [p](x0) = x0,
      
      [s](x0) = 1x0,
      
      [minus](x0, x1) = x0,
      
      [0] = 0
     orientation:
      div#(s(X),s(Y)) = 5X >= 4X = div#(minus(X,Y),s(Y))
      
      minus(X,0()) = X >= X = X
      
      minus(s(X),s(Y)) = 1X >= X = p(minus(X,Y))
      
      p(s(X)) = 1X >= X = X
      
      div(0(),s(Y)) = 0 >= 0 = 0()
      
      div(s(X),s(Y)) = 1X >= 1X = s(div(minus(X,Y),s(Y)))
     problem:
      DPs:
       
      TRS:
       minus(X,0()) -> X
       minus(s(X),s(Y)) -> p(minus(X,Y))
       p(s(X)) -> X
       div(0(),s(Y)) -> 0()
       div(s(X),s(Y)) -> s(div(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)) -> p(minus(X,Y))
     p(s(X)) -> X
     div(0(),s(Y)) -> 0()
     div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))
    Subterm Criterion Processor:
     simple projection:
      pi(minus#) = 1
     problem:
      DPs:
       
      TRS:
       minus(X,0()) -> X
       minus(s(X),s(Y)) -> p(minus(X,Y))
       p(s(X)) -> X
       div(0(),s(Y)) -> 0()
       div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))
     Qed