YES

We show the termination of the TRS R:

  +(x,|0|()) -> x
  +(x,i(x)) -> |0|()
  +(+(x,y),z) -> +(x,+(y,z))
  *(x,+(y,z)) -> +(*(x,y),*(x,z))
  *(+(x,y),z) -> +(*(x,z),*(y,z))

-- SCC decomposition.

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

p1: +#(+(x,y),z) -> +#(x,+(y,z))
p2: +#(+(x,y),z) -> +#(y,z)
p3: *#(x,+(y,z)) -> +#(*(x,y),*(x,z))
p4: *#(x,+(y,z)) -> *#(x,y)
p5: *#(x,+(y,z)) -> *#(x,z)
p6: *#(+(x,y),z) -> +#(*(x,z),*(y,z))
p7: *#(+(x,y),z) -> *#(x,z)
p8: *#(+(x,y),z) -> *#(y,z)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

The estimated dependency graph contains the following SCCs:

  {p4, p5, p7, p8}
  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

The set of usable rules consists of

  (no rules)

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

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

-- Reduction pair.

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

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        +#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2
        +_A(x1,x2) = x1 + x2 + (1,1)
        |0|_A() = (1,1)
        i_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        +#_A(x1,x2) = ((1,1),(1,0)) x1
        +_A(x1,x2) = ((1,0),(1,1)) x1 + (2,1)
        |0|_A() = (1,2)
        i_A(x1) = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        +#_A(x1,x2) = x1
        +_A(x1,x2) = ((1,1),(1,1)) x1 + (2,1)
        |0|_A() = (1,0)
        i_A(x1) = (1,0)
    

The next rules are strictly ordered:

  p1, p2

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