YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      f > a > f#
    
    argument filter:
  
      pi(f#) = [1, 2]
      pi(f) = [1, 2]
      pi(a) = []

The next rules are strictly ordered:

  p1
  r1

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