YES

We show the termination of the TRS R:

  +(x,+(y,z)) -> +(+(x,y),z)
  +(*(x,y),+(x,z)) -> *(x,+(y,z))
  +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)

-- 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)
p3: +#(*(x,y),+(x,z)) -> +#(y,z)
p4: +#(*(x,y),+(*(x,z),u)) -> +#(*(x,+(y,z)),u)
p5: +#(*(x,y),+(*(x,z),u)) -> +#(y,z)

and R consists of:

r1: +(x,+(y,z)) -> +(+(x,y),z)
r2: +(*(x,y),+(x,z)) -> *(x,+(y,z))
r3: +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: +#(x,+(y,z)) -> +#(+(x,y),z)
p2: +#(*(x,y),+(*(x,z),u)) -> +#(y,z)
p3: +#(*(x,y),+(*(x,z),u)) -> +#(*(x,+(y,z)),u)
p4: +#(*(x,y),+(x,z)) -> +#(y,z)
p5: +#(x,+(y,z)) -> +#(x,y)

and R consists of:

r1: +(x,+(y,z)) -> +(+(x,y),z)
r2: +(*(x,y),+(x,z)) -> *(x,+(y,z))
r3: +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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