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)
  Usable Rule Processor:
   DPs:
    f#(+(x,0())) -> f#(x)
    +#(x,+(y,z)) -> +#(x,y)
    +#(x,+(y,z)) -> +#(+(x,y),z)
   TRS:
    +(x,+(y,z)) -> +(+(x,y),z)
   CDG Processor:
    DPs:
     f#(+(x,0())) -> f#(x)
     +#(x,+(y,z)) -> +#(x,y)
     +#(x,+(y,z)) -> +#(+(x,y),z)
    TRS:
     +(x,+(y,z)) -> +(+(x,y),z)
    graph:
     
    Qed