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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        -# > - > neg
      
      argument filter:
    
        pi(-#) = 1
        pi(-) = 1
        pi(neg) = 1
    
    2. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        -#_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,1,0,0)) x1
        -_A(x1,x2) = x1 + (1,1,1,1)
        neg_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0)) x1 + (1,1,1,1)
    

The next rules are strictly ordered:

  p1, p2

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