YES

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

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