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:

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

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^1
    order: standard order
    interpretations:
      *#_A(x1,x2) = x1 + x2
      *_A(x1,x2) = 1
      minus_A(x1) = 0

The next rules are strictly ordered:

  p1

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