YES

Problem:
 f(0(),1(),x) -> f(s(x),x,x)
 f(x,y,s(z)) -> s(f(0(),1(),z))

Proof:
 DP Processor:
  DPs:
   f#(0(),1(),x) -> f#(s(x),x,x)
   f#(x,y,s(z)) -> f#(0(),1(),z)
  TRS:
   f(0(),1(),x) -> f(s(x),x,x)
   f(x,y,s(z)) -> s(f(0(),1(),z))
  Usable Rule Processor:
   DPs:
    f#(0(),1(),x) -> f#(s(x),x,x)
    f#(x,y,s(z)) -> f#(0(),1(),z)
   TRS:
    f5(x,y) -> x
    f5(x,y) -> y
   CDG Processor:
    DPs:
     f#(0(),1(),x) -> f#(s(x),x,x)
     f#(x,y,s(z)) -> f#(0(),1(),z)
    TRS:
     f5(x,y) -> x
     f5(x,y) -> y
    graph:
     f#(0(),1(),x) -> f#(s(x),x,x) -> f#(x,y,s(z)) -> f#(0(),1(),z)
     f#(x,y,s(z)) -> f#(0(),1(),z) -> f#(0(),1(),x) -> f#(s(x),x,x)
     f#(x,y,s(z)) -> f#(0(),1(),z) -> f#(x,y,s(z)) -> f#(0(),1(),z)
    Restore Modifier:
     DPs:
      f#(0(),1(),x) -> f#(s(x),x,x)
      f#(x,y,s(z)) -> f#(0(),1(),z)
     TRS:
      f(0(),1(),x) -> f(s(x),x,x)
      f(x,y,s(z)) -> s(f(0(),1(),z))
     SCC Processor:
      #sccs: 1
      #rules: 2
      #arcs: 3/4
      DPs:
       f#(0(),1(),x) -> f#(s(x),x,x)
       f#(x,y,s(z)) -> f#(0(),1(),z)
      TRS:
       f(0(),1(),x) -> f(s(x),x,x)
       f(x,y,s(z)) -> s(f(0(),1(),z))
      Matrix Interpretation Processor:
       dimension: 1
       interpretation:
        [f#](x0, x1, x2) = x2,
        
        [s](x0) = x0 + 1,
        
        [f](x0, x1, x2) = x2 + 1,
        
        [1] = 0,
        
        [0] = 0
       orientation:
        f#(0(),1(),x) = x >= x = f#(s(x),x,x)
        
        f#(x,y,s(z)) = z + 1 >= z = f#(0(),1(),z)
        
        f(0(),1(),x) = x + 1 >= x + 1 = f(s(x),x,x)
        
        f(x,y,s(z)) = z + 2 >= z + 2 = s(f(0(),1(),z))
       problem:
        DPs:
         f#(0(),1(),x) -> f#(s(x),x,x)
        TRS:
         f(0(),1(),x) -> f(s(x),x,x)
         f(x,y,s(z)) -> s(f(0(),1(),z))
       Matrix Interpretation Processor:
        dimension: 1
        interpretation:
         [f#](x0, x1, x2) = x0 + x2,
         
         [s](x0) = 0,
         
         [f](x0, x1, x2) = 0,
         
         [1] = 0,
         
         [0] = 1
        orientation:
         f#(0(),1(),x) = x + 1 >= x = f#(s(x),x,x)
         
         f(0(),1(),x) = 0 >= 0 = f(s(x),x,x)
         
         f(x,y,s(z)) = 0 >= 0 = s(f(0(),1(),z))
        problem:
         DPs:
          
         TRS:
          f(0(),1(),x) -> f(s(x),x,x)
          f(x,y,s(z)) -> s(f(0(),1(),z))
        Qed