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))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [f#](x0, x1, x2) = [0 0 1]x0 + [1 1 0]x2 + [1],
    
              [1 0 1]     [1]
    [s](x0) = [0 1 0]x0 + [1]
              [0 0 0]     [0],
    
                      [0 0 0]     [0 1 0]     [1]
    [f](x0, x1, x2) = [0 0 1]x0 + [1 1 1]x2 + [0]
                      [0 0 0]     [0 0 0]     [0],
    
          [0]
    [1] = [0]
          [0],
    
          [0]
    [0] = [0]
          [1]
   orientation:
    f#(0(),1(),x) = [1 1 0]x + [2] >= [1 1 0]x + [1] = f#(s(x),x,x)
    
    f#(x,y,s(z)) = [0 0 1]x + [1 1 1]z + [3] >= [1 1 0]z + [2] = f#(0(),1(),z)
    
                   [0 1 0]    [1]    [0 1 0]    [1]              
    f(0(),1(),x) = [1 1 1]x + [1] >= [1 1 1]x + [0] = f(s(x),x,x)
                   [0 0 0]    [0]    [0 0 0]    [0]              
    
                  [0 0 0]    [0 1 0]    [2]    [0 1 0]    [2]                  
    f(x,y,s(z)) = [0 0 1]x + [1 1 1]z + [2] >= [1 1 1]z + [2] = s(f(0(),1(),z))
                  [0 0 0]    [0 0 0]    [0]    [0 0 0]    [0]                  
   problem:
    DPs:
     
    TRS:
     f(0(),1(),x) -> f(s(x),x,x)
     f(x,y,s(z)) -> s(f(0(),1(),z))
   Qed