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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        a > f# > f
      
      argument filter:
    
        pi(f#) = [1, 2]
        pi(f) = []
        pi(a) = []
    
    2. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        f#_A(x1,x2) = ((1,0,0),(1,0,0),(1,0,0)) x1 + ((1,0,0),(0,1,0),(0,1,1)) x2
        f_A(x1,x2) = (1,0,1)
        a_A() = (0,2,0)
    

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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        f > a > f#
      
      argument filter:
    
        pi(f#) = 1
        pi(f) = 2
        pi(a) = []
    
    2. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        f#_A(x1,x2) = x1
        f_A(x1,x2) = ((0,0,0),(1,0,0),(1,1,0)) x2 + (1,0,0)
        a_A() = (2,2,0)
    

The next rules are strictly ordered:

  p1

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