YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1,x2) = x2
        a_A() = 1
        f_A(x1,x2) = x1 + x2
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        f > a > f#
      
      argument filter:
    
        pi(f#) = []
        pi(a) = []
        pi(f) = [2]
    

The next rules are strictly ordered:

  p1

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