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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > * > |1| > i > *#
      
      argument filter:
    
        pi(*#) = [1]
        pi(*) = [1, 2]
        pi(i) = []
        pi(|1|) = []
        pi(|0|) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > |1| > i > * > *#
      
      argument filter:
    
        pi(*#) = 1
        pi(*) = 1
        pi(i) = []
        pi(|1|) = []
        pi(|0|) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        |0| > |1| > i > * > *#
      
      argument filter:
    
        pi(*#) = 1
        pi(*) = 1
        pi(i) = []
        pi(|1|) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

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