YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1,x2) = max{24, x1, x2 - 7}
        f_A(x1,x2) = 25
        a_A = 34
  
    precedence:
    
      f# = f = a
    
    partial status:
  
      pi(f#) = []
      pi(f) = []
      pi(a) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1,x2) = max{0, x1 - 3, x2 - 28}
        f_A(x1,x2) = max{2, x2 - 15}
        a_A = 26
  
    precedence:
    
      a > f# = f
    
    partial status:
  
      pi(f#) = []
      pi(f) = []
      pi(a) = []

The next rules are strictly ordered:

  p1

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