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))
  Subterm Criterion Processor:
   simple projection:
    pi(*#) = 0
   problem:
    DPs:
     
    TRS:
     *(i(x),x) -> 1()
     *(1(),y) -> y
     *(x,0()) -> 0()
     *(*(x,y),z) -> *(x,*(y,z))
   Qed