YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: a(f(),a(g(),a(f(),x))) -> a(f(),a(g(),a(g(),a(f(),x))))
r2: a(g(),a(f(),a(g(),x))) -> a(g(),a(f(),a(f(),a(g(),x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

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

and R consists of:

r1: a(f(),a(g(),a(f(),x))) -> a(f(),a(g(),a(g(),a(f(),x))))
r2: a(g(),a(f(),a(g(),x))) -> a(g(),a(f(),a(f(),a(g(),x))))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    a#_A(x1,x2) = max{x1 + 1, x2 + 3}
    f_A = 0
    a_A(x1,x2) = max{x1 + 4, x2 - 1}
    g_A = 5

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a(f(),a(g(),a(f(),x))) -> a(f(),a(g(),a(g(),a(f(),x))))
r2: a(g(),a(f(),a(g(),x))) -> a(g(),a(f(),a(f(),a(g(),x))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

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

and R consists of:

r1: a(f(),a(g(),a(f(),x))) -> a(f(),a(g(),a(g(),a(f(),x))))
r2: a(g(),a(f(),a(g(),x))) -> a(g(),a(f(),a(f(),a(g(),x))))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    a#_A(x1,x2) = max{x1 + 10, x2 + 5}
    f_A = 0
    a_A(x1,x2) = x1 + 8
    g_A = 5

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a(f(),a(g(),a(f(),x))) -> a(f(),a(g(),a(g(),a(f(),x))))
r2: a(g(),a(f(),a(g(),x))) -> a(g(),a(f(),a(f(),a(g(),x))))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: a(f(),a(g(),a(f(),x))) -> a(f(),a(g(),a(g(),a(f(),x))))
r2: a(g(),a(f(),a(g(),x))) -> a(g(),a(f(),a(f(),a(g(),x))))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    a#_A(x1,x2) = max{x1, x2 + 23}
    f_A = 19
    a_A(x1,x2) = max{1, x1 - 5, x2 - 4}
    g_A = 0

The next rules are strictly ordered:

  p1

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