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))
  EDG 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))
   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)
   Subterm Criterion Processor:
    simple projection:
     pi(f#) = 2
    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