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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        *#_A(x1,x2) = max{x1 + 5, x2 + 2}
        *_A(x1,x2) = max{x1 + 3, x2}
        i_A(x1) = x1 + 3
        |1|_A = 2
        |0|_A = 2
  
    precedence:
    
      *# = * = i = |1| = |0|
    
    partial status:
  
      pi(*#) = [1]
      pi(*) = [1, 2]
      pi(i) = [1]
      pi(|1|) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: *#(*(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}


-- Reduction pair.

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

p1: *#(*(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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        *#_A(x1,x2) = max{x1 + 1, x2 + 1}
        *_A(x1,x2) = x2
  
    precedence:
    
      *# = *
    
    partial status:
  
      pi(*#) = [1]
      pi(*) = [2]

The next rules are strictly ordered:

  p1

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