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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            *#_A(x1,x2) = ((1,0),(1,1)) x1 + (1,2)
            *_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,1)
            i_A(x1) = ((1,0),(1,1)) x1 + (1,1)
            |1|_A() = (2,4)
            |0|_A() = (1,1)
    
      precedence:
      
        *# = * = i = |1| = |0|
      
      partial status:
    
        pi(*#) = []
        pi(*) = []
        pi(i) = []
        pi(|1|) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            *#_A(x1,x2) = x1 + (1,1)
            *_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (2,2)
            i_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            |1|_A() = (1,1)
            |0|_A() = (1,1)
    
      precedence:
      
        * > *# = i = |0| > |1|
      
      partial status:
    
        pi(*#) = [1]
        pi(*) = []
        pi(i) = [1]
        pi(|1|) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: *#(*(x,y),z) -> *#(x,*(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) -> *#(x,*(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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            *#_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1)
            *_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1)
            i_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            |1|_A() = (1,1)
            |0|_A() = (1,1)
    
      precedence:
      
        *# = * = i = |1| = |0|
      
      partial status:
    
        pi(*#) = []
        pi(*) = []
        pi(i) = []
        pi(|1|) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            *#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (1,1)
            *_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,3)
            i_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            |1|_A() = (1,1)
            |0|_A() = (1,1)
    
      precedence:
      
        * > *# = |1| = |0| > i
      
      partial status:
    
        pi(*#) = [2]
        pi(*) = [2]
        pi(i) = []
        pi(|1|) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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