YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1,x2) = max{8, x1, x2 - 5}
        f_A(x1,x2) = max{4, x1 - 4, x2 - 3}
        a_A = 12
  
    precedence:
    
      f > f# = a
    
    partial status:
  
      pi(f#) = [1]
      pi(f) = []
      pi(a) = []

The next rules are strictly ordered:

  p1

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