YES

Problem:
 p(0()) -> 0()
 p(s(x)) -> x
 minus(x,0()) -> x
 minus(x,s(y)) -> minus(p(x),y)

Proof:
 DP Processor:
  DPs:
   minus#(x,s(y)) -> p#(x)
   minus#(x,s(y)) -> minus#(p(x),y)
  TRS:
   p(0()) -> 0()
   p(s(x)) -> x
   minus(x,0()) -> x
   minus(x,s(y)) -> minus(p(x),y)
  Usable Rule Processor:
   DPs:
    minus#(x,s(y)) -> p#(x)
    minus#(x,s(y)) -> minus#(p(x),y)
   TRS:
    p(0()) -> 0()
    p(s(x)) -> x
   CDG Processor:
    DPs:
     minus#(x,s(y)) -> p#(x)
     minus#(x,s(y)) -> minus#(p(x),y)
    TRS:
     p(0()) -> 0()
     p(s(x)) -> x
    graph:
     
    Qed