YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

r1: f(f(x,a()),y) -> f(f(a(),y),f(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^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = ((1,1),(0,0)) x1 + x2
        f_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (0,1)
        a_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,1),(0,1)) x2
        f_A(x1,x2) = ((0,1),(0,0)) x2 + (1,1)
        a_A() = (1,2)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = (0,0)
        f_A(x1,x2) = (1,1)
        a_A() = (0,0)
    

The next rules are strictly ordered:

  p1

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