YES

We show the termination of the TRS R:

  *(x,*(y,z)) -> *(*(x,y),z)
  *(x,x) -> x

-- SCC decomposition.

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

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

and R consists of:

r1: *(x,*(y,z)) -> *(*(x,y),z)
r2: *(x,x) -> x

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)) -> *#(x,y)

and R consists of:

r1: *(x,*(y,z)) -> *(*(x,y),z)
r2: *(x,x) -> x

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = max{0, x2 - 4}
          *_A(x1,x2) = max{5, x1, x2 + 3}
    
      precedence:
      
        *# = *
      
      partial status:
    
        pi(*#) = []
        pi(*) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = 0
          *_A(x1,x2) = x2 + 2
    
      precedence:
      
        * > *#
      
      partial status:
    
        pi(*#) = []
        pi(*) = []
    

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)) -> *#(x,y)

and R consists of:

r1: *(x,*(y,z)) -> *(*(x,y),z)
r2: *(x,x) -> x

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)

and R consists of:

r1: *(x,*(y,z)) -> *(*(x,y),z)
r2: *(x,x) -> x

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = max{0, x2 - 2}
          *_A(x1,x2) = max{x1 + 1, x2 + 3}
    
      precedence:
      
        *# = *
      
      partial status:
    
        pi(*#) = []
        pi(*) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = 0
          *_A(x1,x2) = max{x1 + 1, x2}
    
      precedence:
      
        *# = *
      
      partial status:
    
        pi(*#) = []
        pi(*) = [2]
    

The next rules are strictly ordered:

  p1

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