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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        *#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,0,0,1)) x1 + x2
        *_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(1,1,1,0),(0,0,0,1)) x1 + ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,0,0)) x2 + (1,2,2,1)
        minus_A(x1) = x1 + (1,0,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        minus > *# > *
      
      argument filter:
    
        pi(*#) = [1, 2]
        pi(*) = []
        pi(minus) = []
    

The next rules are strictly ordered:

  p1, p2

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