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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        *#_A(x1,x2) = x2 + 4
        *_A(x1,x2) = max{x1, x2 + 5}
  
    precedence:
    
      *# = *
    
    partial status:
  
      pi(*#) = []
      pi(*) = [2]

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:

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

The next rules are strictly ordered:

  p1

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