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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        *#_A(x1,x2) = ((1,1),(0,0)) x1 + x2
        *_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        *#_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,0),(1,1)) x2
        *_A(x1,x2) = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        *#_A(x1,x2) = ((0,1),(0,0)) x1 + x2
        *_A(x1,x2) = (2,1)
    

The next rules are strictly ordered:

  p1, p2

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