YES

Problem:
 f(s(x1)) -> s(s(f(p(s(x1)))))
 f(0(x1)) -> 0(x1)
 p(s(x1)) -> x1

Proof:
 DP Processor:
  DPs:
   f#(s(x1)) -> p#(s(x1))
   f#(s(x1)) -> f#(p(s(x1)))
  TRS:
   f(s(x1)) -> s(s(f(p(s(x1)))))
   f(0(x1)) -> 0(x1)
   p(s(x1)) -> x1
  TDG Processor:
   DPs:
    f#(s(x1)) -> p#(s(x1))
    f#(s(x1)) -> f#(p(s(x1)))
   TRS:
    f(s(x1)) -> s(s(f(p(s(x1)))))
    f(0(x1)) -> 0(x1)
    p(s(x1)) -> x1
   graph:
    f#(s(x1)) -> f#(p(s(x1))) -> f#(s(x1)) -> f#(p(s(x1)))
    f#(s(x1)) -> f#(p(s(x1))) -> f#(s(x1)) -> p#(s(x1))
   CDG Processor:
    DPs:
     f#(s(x1)) -> p#(s(x1))
     f#(s(x1)) -> f#(p(s(x1)))
    TRS:
     f(s(x1)) -> s(s(f(p(s(x1)))))
     f(0(x1)) -> 0(x1)
     p(s(x1)) -> x1
    graph:
     
    Qed