YES

Problem:
 *(i(x),x) -> 1()
 *(1(),y) -> y
 *(x,0()) -> 0()
 *(*(x,y),z) -> *(x,*(y,z))

Proof:
 DP Processor:
  DPs:
   *#(*(x,y),z) -> *#(y,z)
   *#(*(x,y),z) -> *#(x,*(y,z))
  TRS:
   *(i(x),x) -> 1()
   *(1(),y) -> y
   *(x,0()) -> 0()
   *(*(x,y),z) -> *(x,*(y,z))
  CDG Processor:
   DPs:
    *#(*(x,y),z) -> *#(y,z)
    *#(*(x,y),z) -> *#(x,*(y,z))
   TRS:
    *(i(x),x) -> 1()
    *(1(),y) -> y
    *(x,0()) -> 0()
    *(*(x,y),z) -> *(x,*(y,z))
   graph:
    
   Qed