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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        *# > + > minus
      
      argument filter:
    
        pi(*#) = 2
        pi(minus) = 1
        pi(+) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        *# > + > minus
      
      argument filter:
    
        pi(*#) = [2]
        pi(minus) = 1
        pi(+) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        *# > + > minus
      
      argument filter:
    
        pi(*#) = 2
        pi(minus) = [1]
        pi(+) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        + > minus#
      
      argument filter:
    
        pi(minus#) = 1
        pi(+) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        + > minus#
      
      argument filter:
    
        pi(minus#) = 1
        pi(+) = [1]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        + > minus#
      
      argument filter:
    
        pi(minus#) = [1]
        pi(+) = [1]
    

The next rules are strictly ordered:

  p1, p2

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