YES

We show the termination of the TRS R:

  *(x,*(minus(y),y)) -> *(minus(*(y,y)),x)

-- SCC decomposition.

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

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

and R consists of:

r1: *(x,*(minus(y),y)) -> *(minus(*(y,y)),x)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: *(x,*(minus(y),y)) -> *(minus(*(y,y)),x)

The set of usable rules consists of

  r1

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1, p2
  r1

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