YES

We show the termination of the TRS R:

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

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: *#(*(x,y),z) -> *#(x,*(y,z))
p2: *#(*(x,y),z) -> *#(y,z)

and R consists of:

r1: *(i(x),x) -> |1|()
r2: *(|1|(),y) -> y
r3: *(x,|0|()) -> |0|()
r4: *(*(x,y),z) -> *(x,*(y,z))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: *#(*(x,y),z) -> *#(x,*(y,z))
p2: *#(*(x,y),z) -> *#(y,z)

and R consists of:

r1: *(i(x),x) -> |1|()
r2: *(|1|(),y) -> y
r3: *(x,|0|()) -> |0|()
r4: *(*(x,y),z) -> *(x,*(y,z))

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      *#_A(x1,x2) = ((0,1),(1,1)) x1 + x2
      *_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,1)
      i_A(x1) = ((1,1),(1,1)) x1 + (1,1)
      |1|_A() = (1,1)
      |0|_A() = (0,0)

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.