YES

We show the termination of the TRS R:

  f(f(x,a()),y) -> f(f(a(),y),f(a(),x))

-- SCC decomposition.

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

p1: f#(f(x,a()),y) -> f#(f(a(),y),f(a(),x))
p2: f#(f(x,a()),y) -> f#(a(),y)
p3: f#(f(x,a()),y) -> f#(a(),x)

and R consists of:

r1: f(f(x,a()),y) -> f(f(a(),y),f(a(),x))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(f(x,a()),y) -> f#(f(a(),y),f(a(),x))

and R consists of:

r1: f(f(x,a()),y) -> f(f(a(),y),f(a(),x))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      f#_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2
      f_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x2 + (1,1,1,1)
      a_A() = (2,1,1,1)

The next rules are strictly ordered:

  p1

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