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)))
  KBO Processor:
   argument filtering:
    pi(0) = []
    pi(minus) = 0
    pi(s) = [0]
    pi(p) = 0
    pi(div) = 0
    pi(minus#) = 0
    pi(p#) = 0
    pi(div#) = 0
   weight function:
    w0 = 1
    w(div#) = w(minus#) = w(p) = w(0) = 1
    w(p#) = w(div) = w(s) = w(minus) = 0
   precedence:
    div# ~ p# ~ minus# ~ div ~ p ~ s ~ minus ~ 0
   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