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