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