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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        a > f > f#
      
      argument filter:
    
        pi(f#) = [1, 2]
        pi(f) = [1, 2]
        pi(a) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        f > a > f#
      
      argument filter:
    
        pi(f#) = [1, 2]
        pi(f) = [2]
        pi(a) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        f > f# > a
      
      argument filter:
    
        pi(f#) = []
        pi(f) = 2
        pi(a) = []
    

The next rules are strictly ordered:

  p1

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