YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: +#(*(x,y),*(a(),y)) -> *#(+(x,a()),y)
p2: +#(*(x,y),*(a(),y)) -> +#(x,a())
p3: *#(*(x,y),z) -> *#(x,*(y,z))
p4: *#(*(x,y),z) -> *#(y,z)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p3, p4}


-- 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: +(*(x,y),*(a(),y)) -> *(+(x,a()),y)
r2: *(*(x,y),z) -> *(x,*(y,z))

The set of usable rules consists of

  r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        *# > *
      
      argument filter:
    
        pi(*#) = [1, 2]
        pi(*) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        * > *#
      
      argument filter:
    
        pi(*#) = [1, 2]
        pi(*) = 2
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        * > *#
      
      argument filter:
    
        pi(*#) = [1, 2]
        pi(*) = 2
    

The next rules are strictly ordered:

  p1, p2

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