YES

We show the termination of the TRS R:

  *(x,+(y,z)) -> +(*(x,y),*(x,z))
  *(+(x,y),z) -> +(*(x,z),*(y,z))
  *(x,|1|()) -> x
  *(|1|(),y) -> y

-- SCC decomposition.

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

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

and R consists of:

r1: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r2: *(+(x,y),z) -> +(*(x,z),*(y,z))
r3: *(x,|1|()) -> x
r4: *(|1|(),y) -> y

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

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

and R consists of:

r1: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r2: *(+(x,y),z) -> +(*(x,z),*(y,z))
r3: *(x,|1|()) -> x
r4: *(|1|(),y) -> y

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2, p3, p4

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