YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: f#(x,f(a(),y)) -> f#(f(y,f(f(a(),a()),a())),x)
p2: f#(x,f(a(),y)) -> f#(y,f(f(a(),a()),a()))
p3: f#(x,f(a(),y)) -> f#(f(a(),a()),a())
p4: f#(x,f(a(),y)) -> f#(a(),a())

and R consists of:

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

and R consists of:

r1: f(x,f(a(),y)) -> f(f(y,f(f(a(),a()),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),(1,0,0,0),(0,1,0,0),(0,1,1,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,1,0)) 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),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (1,1,0,1)
      a_A() = (2,1,1,0)

The next rules are strictly ordered:

  p1

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