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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            -#_A(x1,x2) = x2 + (1,2)
            -_A(x1,x2) = ((1,0),(0,0)) x2 + (1,1)
            neg_A(x1) = ((1,0),(1,1)) x1 + (3,4)
    
      precedence:
      
        neg > -# = -
      
      partial status:
    
        pi(-#) = [2]
        pi(-) = []
        pi(neg) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            -#_A(x1,x2) = (0,0)
            -_A(x1,x2) = (1,1)
            neg_A(x1) = ((1,0),(1,1)) x1 + (2,2)
    
      precedence:
      
        -# = - = neg
      
      partial status:
    
        pi(-#) = []
        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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            -#_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (1,3)
            -_A(x1,x2) = x1 + x2 + (2,2)
            neg_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        -# = - = neg
      
      partial status:
    
        pi(-#) = []
        pi(-) = []
        pi(neg) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            -#_A(x1,x2) = ((0,0),(1,0)) x2 + (1,3)
            -_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2
            neg_A(x1) = (2,0)
    
      precedence:
      
        neg > - > -#
      
      partial status:
    
        pi(-#) = []
        pi(-) = []
        pi(neg) = []
    

The next rules are strictly ordered:

  p1

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