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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        *#_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2
        +_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        *#_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2
        +_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3, p4
  r1, r2, r3, r4

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