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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          -#_A(x1,x2) = x2 + 1
          -_A(x1,x2) = x2 + 1
          neg_A(x1) = x1 + 3
  
    precedence:
    
      -# = - = neg
    
    partial status:
  
      pi(-#) = [2]
      pi(-) = []
      pi(neg) = [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: -#(-(neg(x),neg(x)),-(neg(y),neg(y))) -> -#(-(x,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}


-- 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))

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          -#_A(x1,x2) = x1 + 1
          -_A(x1,x2) = x1 + 3
          neg_A(x1) = x1 + 5
  
    precedence:
    
      -# = - = neg
    
    partial status:
  
      pi(-#) = []
      pi(-) = [1]
      pi(neg) = [1]

The next rules are strictly ordered:

  p1

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