YES

Problem:
 g(x,y) -> x
 g(x,y) -> y
 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:
   g(x,y) -> x
   g(x,y) -> y
   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:
    
   Matrix Interpretation Processor: dim=4
    
    usable rules:
     
    interpretation:
     [f#](x0, x1, x2) = [0 0 1 0]x0 + [0 1 0 1]x2,
     
               [0 0 0 0]     [0]
               [1 0 0 0]     [1]
     [s](x0) = [0 0 0 0]x0 + [0]
               [0 1 0 1]     [1],
     
           [0]
           [0]
     [1] = [0]
           [0],
     
           [0]
           [0]
     [0] = [1]
           [0]
    orientation:
     f#(0(),1(),x) = [0 1 0 1]x + [1] >= [0 1 0 1]x = f#(s(x),x,x)
     
     f#(x,y,s(z)) = [0 0 1 0]x + [1 1 0 1]z + [2] >= [0 1 0 1]z + [1] = f#(0(),1(),z)
    problem:
     DPs:
      
     TRS:
      
    Qed