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 monotone reduction pair:

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

The next rules are strictly ordered:

  p2
  r1

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,z))

and R consists of:

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

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,z))

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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