YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(x),y) -> f#(x,y)
   f#(s(x),y) -> f#(f(x,y),y)
  TRS:
   f(0(),y) -> 0()
   f(s(x),y) -> f(f(x,y),y)
  CDG Processor:
   DPs:
    f#(s(x),y) -> f#(x,y)
    f#(s(x),y) -> f#(f(x,y),y)
   TRS:
    f(0(),y) -> 0()
    f(s(x),y) -> f(f(x,y),y)
   graph:
    f#(s(x),y) -> f#(x,y) -> f#(s(x),y) -> f#(x,y)
    f#(s(x),y) -> f#(x,y) -> f#(s(x),y) -> f#(f(x,y),y)
   Restore Modifier:
    DPs:
     f#(s(x),y) -> f#(x,y)
     f#(s(x),y) -> f#(f(x,y),y)
    TRS:
     f(0(),y) -> 0()
     f(s(x),y) -> f(f(x,y),y)
    SCC Processor:
     #sccs: 1
     #rules: 1
     #arcs: 2/4
     DPs:
      f#(s(x),y) -> f#(x,y)
     TRS:
      f(0(),y) -> 0()
      f(s(x),y) -> f(f(x,y),y)
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [f#](x0, x1) = x0 + 1,
       
       [s](x0) = x0 + 1,
       
       [f](x0, x1) = 0,
       
       [0] = 0
      orientation:
       f#(s(x),y) = x + 2 >= x + 1 = f#(x,y)
       
       f(0(),y) = 0 >= 0 = 0()
       
       f(s(x),y) = 0 >= 0 = f(f(x,y),y)
      problem:
       DPs:
        
       TRS:
        f(0(),y) -> 0()
        f(s(x),y) -> f(f(x,y),y)
      Qed