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. lexicographic path order with precedence:
    
      precedence:
      
        - > -# > neg
      
      argument filter:
    
        pi(-#) = [1]
        pi(-) = [1]
        pi(neg) = [1]
    

The next rules are strictly ordered:

  p1, p2

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