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)
   KBO Processor:
    argument filtering:
     pi(0) = []
     pi(+) = [0,1]
     pi(f#) = 0
     pi(+#) = 1
    usable rules:
     
    weight function:
     w0 = 1
     w(+#) = w(f#) = w(0) = 1
     w(+) = 0
    precedence:
     +# ~ f# ~ + ~ 0
    problem:
     DPs:
      
     TRS:
      +(x,+(y,z)) -> +(+(x,y),z)
    Qed