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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      *#_A(x1,x2) = max{x1 + 1, x2}
      *_A(x1,x2) = max{9, x1 + 7, x2 + 7}
      minus_A(x1) = max{1, x1 - 4}
    
    2. max/plus interpretations on natural numbers:
    
      *#_A(x1,x2) = 0
      *_A(x1,x2) = 0
      minus_A(x1) = 0
    

The next rules are strictly ordered:

  p1, p2

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