YES

We show the termination of the TRS R:

  +(x,|0|()) -> x
  +(minus(x),x) -> |0|()
  minus(|0|()) -> |0|()
  minus(minus(x)) -> x
  minus(+(x,y)) -> +(minus(y),minus(x))
  *(x,|1|()) -> x
  *(x,|0|()) -> |0|()
  *(x,+(y,z)) -> +(*(x,y),*(x,z))
  *(x,minus(y)) -> minus(*(x,y))

-- SCC decomposition.

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

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(minus(x),x) -> |0|()
r3: minus(|0|()) -> |0|()
r4: minus(minus(x)) -> x
r5: minus(+(x,y)) -> +(minus(y),minus(x))
r6: *(x,|1|()) -> x
r7: *(x,|0|()) -> |0|()
r8: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r9: *(x,minus(y)) -> minus(*(x,y))

The estimated dependency graph contains the following SCCs:

  {p5, p6, p8}
  {p2, p3}


-- Reduction pair.

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

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(minus(x),x) -> |0|()
r3: minus(|0|()) -> |0|()
r4: minus(minus(x)) -> x
r5: minus(+(x,y)) -> +(minus(y),minus(x))
r6: *(x,|1|()) -> x
r7: *(x,|0|()) -> |0|()
r8: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r9: *(x,minus(y)) -> minus(*(x,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) = x1 + ((1,1),(1,0)) x2
        minus_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        +_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) = x1 + ((1,1),(1,1)) x2
        minus_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        +_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3
  r1, r2, r3, r4, r5, r6, r7, r8, r9

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: minus#(+(x,y)) -> minus#(y)
p2: minus#(+(x,y)) -> minus#(x)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(minus(x),x) -> |0|()
r3: minus(|0|()) -> |0|()
r4: minus(minus(x)) -> x
r5: minus(+(x,y)) -> +(minus(y),minus(x))
r6: *(x,|1|()) -> x
r7: *(x,|0|()) -> |0|()
r8: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r9: *(x,minus(y)) -> minus(*(x,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:
        minus#_A(x1) = ((1,1),(1,1)) x1
        +_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:
        minus#_A(x1) = ((1,1),(1,0)) x1
        +_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,0)) x2 + (1,1)
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9

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