YES

Problem:
 +(x,+(y,z)) -> +(+(x,y),z)
 +(*(x,y),+(x,z)) -> *(x,+(y,z))
 +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)

Proof:
 DP Processor:
  DPs:
   +#(x,+(y,z)) -> +#(x,y)
   +#(x,+(y,z)) -> +#(+(x,y),z)
   +#(*(x,y),+(x,z)) -> +#(y,z)
   +#(*(x,y),+(*(x,z),u)) -> +#(y,z)
   +#(*(x,y),+(*(x,z),u)) -> +#(*(x,+(y,z)),u)
  TRS:
   +(x,+(y,z)) -> +(+(x,y),z)
   +(*(x,y),+(x,z)) -> *(x,+(y,z))
   +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)
  Subterm Criterion Processor:
   simple projection:
    pi(+#) = 1
   problem:
    DPs:
     
    TRS:
     +(x,+(y,z)) -> +(+(x,y),z)
     +(*(x,y),+(x,z)) -> *(x,+(y,z))
     +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)
   Qed