YES

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

Proof:
 DP Processor:
  DPs:
   f#(+(x,0())) -> f#(x)
   +#(x,+(y,z)) -> +#(x,y)
   +#(x,+(y,z)) -> +#(+(x,y),z)
  TRS:
   f(+(x,0())) -> f(x)
   +(x,+(y,z)) -> +(+(x,y),z)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [+#](x0, x1) = x1,
    
    [f#](x0) = 4x0,
    
    [f](x0) = 2x0 + 4,
    
    [+](x0, x1) = x0 + 4x1 + 6,
    
    [0] = 0
   orientation:
    f#(+(x,0())) = 4x + 24 >= 4x = f#(x)
    
    +#(x,+(y,z)) = y + 4z + 6 >= y = +#(x,y)
    
    +#(x,+(y,z)) = y + 4z + 6 >= z = +#(+(x,y),z)
    
    f(+(x,0())) = 2x + 16 >= 2x + 4 = f(x)
    
    +(x,+(y,z)) = x + 4y + 16z + 30 >= x + 4y + 4z + 12 = +(+(x,y),z)
   problem:
    DPs:
     
    TRS:
     f(+(x,0())) -> f(x)
     +(x,+(y,z)) -> +(+(x,y),z)
   Qed