YES

We show the termination of the TRS R:

  +(*(x,y),*(x,z)) -> *(x,+(y,z))
  +(+(x,y),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),*(x,z)) -> +#(y,z)
p2: +#(+(x,y),z) -> +#(x,+(y,z))
p3: +#(+(x,y),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),*(x,z)) -> *(x,+(y,z))
r2: +(+(x,y),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, p5}


-- Reduction pair.

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

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

and R consists of:

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

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

The next rules are strictly ordered:

  p1, p2, p3, p4

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