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:
    f7(x,y) -> x
    f7(x,y) -> y
   Restore Modifier:
    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))
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 4/4
     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))
     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,
       
       [g](x0, x1) = x0 + x1
      orientation:
       f#(0(),1(),x) = x >= x = f#(s(x),x,x)
       
       f#(x,y,s(z)) = z + 1 >= z = f#(0(),1(),z)
       
       g(x,y) = x + y >= x = x
       
       g(x,y) = x + y >= y = y
       
       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:
        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))
      Matrix Interpretation Processor:
       dimension: 1
       interpretation:
        [f#](x0, x1, x2) = x0,
        
        [s](x0) = 0,
        
        [f](x0, x1, x2) = 0,
        
        [1] = 0,
        
        [0] = 1,
        
        [g](x0, x1) = x0 + x1
       orientation:
        f#(0(),1(),x) = 1 >= 0 = f#(s(x),x,x)
        
        g(x,y) = x + y >= x = x
        
        g(x,y) = x + y >= y = y
        
        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:
         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))
       Qed