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:

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

The next rules are strictly ordered:

  p2

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)
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),z)

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:

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

The next rules are strictly ordered:

  p1

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