YES

Problem:
 f(s(x),y) -> f(x,f(x,y))
 f(0(),y) -> c(y,y)

Proof:
 DP Processor:
  DPs:
   f#(s(x),y) -> f#(x,y)
   f#(s(x),y) -> f#(x,f(x,y))
  TRS:
   f(s(x),y) -> f(x,f(x,y))
   f(0(),y) -> c(y,y)
  Subterm Criterion Processor:
   simple projection:
    pi(f#) = 0
   problem:
    DPs:
     
    TRS:
     f(s(x),y) -> f(x,f(x,y))
     f(0(),y) -> c(y,y)
   Qed