YES

We show the termination of the TRS R:

  -(-(neg(x),neg(x)),-(neg(y),neg(y))) -> -(-(x,y),-(x,y))

-- SCC decomposition.

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

p1: -#(-(neg(x),neg(x)),-(neg(y),neg(y))) -> -#(-(x,y),-(x,y))
p2: -#(-(neg(x),neg(x)),-(neg(y),neg(y))) -> -#(x,y)

and R consists of:

r1: -(-(neg(x),neg(x)),-(neg(y),neg(y))) -> -(-(x,y),-(x,y))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: -#(-(neg(x),neg(x)),-(neg(y),neg(y))) -> -#(-(x,y),-(x,y))
p2: -#(-(neg(x),neg(x)),-(neg(y),neg(y))) -> -#(x,y)

and R consists of:

r1: -(-(neg(x),neg(x)),-(neg(y),neg(y))) -> -(-(x,y),-(x,y))

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
      -_A(x1,x2) = x1 + 1
      neg_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1, p2

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