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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        *#_A(x1,x2) = x2
        *_A(x1,x2) = x1 + x2 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        * > *#
      
      argument filter:
    
        pi(*#) = [2]
        pi(*) = [2]
    

The next rules are strictly ordered:

  p1, p2

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