YES

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

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