YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

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

and R consists of:

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

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) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,1,1,1)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,0,0),(0,1,1,0)) x2
      f_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x2 + (1,0,1,1)
      a_A() = (2,1,1,1)

The next rules are strictly ordered:

  p1

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