YES

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

Proof:
 Polynomial Interpretation Processor:
  dimension: 1
  interpretation:
   [0] = 0,
   
   [1] = 0,
   
   [*](x0, x1) = 5x0 + x1 + 1,
   
   [i](x0) = 4x0
  orientation:
   *(i(x),x) = 21x + 1 >= 0 = 1()
   
   *(1(),y) = y + 1 >= y = y
   
   *(x,0()) = 5x + 1 >= 0 = 0()
   
   *(*(x,y),z) = 25x + 5y + z + 6 >= 5x + 5y + z + 2 = *(x,*(y,z))
  problem:
   
  Qed