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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      *#_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2
      *_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(1,1,0,0)) x1 + ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,1,0,0)) x2 + (1,2,1,1)
      minus_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,0,0,0)) x1 + (1,0,1,1)

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      *#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,1,0)) x2
      *_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,1,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,0,0)) x2 + (1,0,1,1)
      minus_A(x1) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,0,1,1)

The next rules are strictly ordered:

  p1

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