YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        f#_A(x1,x2) = x1 + ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,0,0,1)) x2
        f_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,1,1)) x1 + (1,1,0,0)
        a_A() = (1,0,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        a > f > f#
      
      argument filter:
    
        pi(f#) = []
        pi(f) = [1]
        pi(a) = []
    

The next rules are strictly ordered:

  p1

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