YES

We show the termination of the TRS R:

  +(-(x,y),z) -> -(+(x,z),y)
  -(+(x,y),y) -> x

-- SCC decomposition.

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

p1: +#(-(x,y),z) -> -#(+(x,z),y)
p2: +#(-(x,y),z) -> +#(x,z)

and R consists of:

r1: +(-(x,y),z) -> -(+(x,z),y)
r2: -(+(x,y),y) -> x

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: +#(-(x,y),z) -> +#(x,z)

and R consists of:

r1: +(-(x,y),z) -> -(+(x,z),y)
r2: -(+(x,y),y) -> x

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        +#_A(x1,x2) = ((1,1),(1,1)) x1 + x2
        -_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        +#_A(x1,x2) = ((0,1),(1,0)) x1 + x2
        -_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,1),(0,1)) x2 + (1,1)
    

The next rules are strictly ordered:

  p1

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