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 monotone reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      +#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x2
      -_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + (1,1,1,1)

The next rules are strictly ordered:

  p1
  r1, r2

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