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)
  TDG 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)
   graph:
    +#(x,+(y,z)) -> +#(+(x,y),z) -> +#(x,+(y,z)) -> +#(+(x,y),z)
    +#(x,+(y,z)) -> +#(+(x,y),z) -> +#(x,+(y,z)) -> +#(x,y)
    +#(x,+(y,z)) -> +#(x,y) -> +#(x,+(y,z)) -> +#(+(x,y),z)
    +#(x,+(y,z)) -> +#(x,y) -> +#(x,+(y,z)) -> +#(x,y)
    f#(+(x,0())) -> f#(x) -> f#(+(x,0())) -> f#(x)
   SCC Processor:
    #sccs: 2
    #rules: 3
    #arcs: 5/9
    DPs:
     f#(+(x,0())) -> f#(x)
    TRS:
     f(+(x,0())) -> f(x)
     +(x,+(y,z)) -> +(+(x,y),z)
    Subterm Criterion Processor:
     simple projection:
      pi(f#) = 0
     problem:
      DPs:
       
      TRS:
       f(+(x,0())) -> f(x)
       +(x,+(y,z)) -> +(+(x,y),z)
     Qed
    
    DPs:
     +#(x,+(y,z)) -> +#(+(x,y),z)
     +#(x,+(y,z)) -> +#(x,y)
    TRS:
     f(+(x,0())) -> f(x)
     +(x,+(y,z)) -> +(+(x,y),z)
    Subterm Criterion Processor:
     simple projection:
      pi(+#) = 1
     problem:
      DPs:
       
      TRS:
       f(+(x,0())) -> f(x)
       +(x,+(y,z)) -> +(+(x,y),z)
     Qed