YES

Problem:
 bin(x,0()) -> s(0())
 bin(0(),s(y)) -> 0()
 bin(s(x),s(y)) -> +(bin(x,s(y)),bin(x,y))

Proof:
 DP Processor:
  DPs:
   bin#(s(x),s(y)) -> bin#(x,y)
   bin#(s(x),s(y)) -> bin#(x,s(y))
  TRS:
   bin(x,0()) -> s(0())
   bin(0(),s(y)) -> 0()
   bin(s(x),s(y)) -> +(bin(x,s(y)),bin(x,y))
  KBO Processor:
   argument filtering:
    pi(0) = []
    pi(bin) = [0,1]
    pi(s) = [0]
    pi(+) = [0]
    pi(bin#) = 0
   weight function:
    w0 = 1
    w(bin#) = w(+) = w(s) = w(bin) = w(0) = 1
   precedence:
    bin# ~ s ~ bin > + ~ 0
   problem:
    DPs:
     
    TRS:
     bin(x,0()) -> s(0())
     bin(0(),s(y)) -> 0()
     bin(s(x),s(y)) -> +(bin(x,s(y)),bin(x,y))
   Qed