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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2) = max{1, x1 - 23, x2 - 8}
        f_A = 12
        a_A(x1,x2) = max{12, x1 + 8, x2 - 1}
        g_A = 8
  
    precedence:
    
      a# = f = a = g
    
    partial status:
  
      pi(a#) = []
      pi(f) = []
      pi(a) = []
      pi(g) = []

The next rules are strictly ordered:

  p1

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

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2) = max{1, x1 - 20, x2 - 1}
        f_A = 0
        a_A(x1,x2) = max{5, x1 - 4, x2 - 1}
        g_A = 13
  
    precedence:
    
      a = g > a# = f
    
    partial status:
  
      pi(a#) = []
      pi(f) = []
      pi(a) = []
      pi(g) = []

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


-- Reduction pair.

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

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

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2) = max{x1 + 7, x2 - 15}
        f_A = 12
        a_A(x1,x2) = max{28, x1 + 27}
        g_A = 2
  
    precedence:
    
      f > a# = a = g
    
    partial status:
  
      pi(a#) = []
      pi(f) = []
      pi(a) = []
      pi(g) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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:

  (no SCCs)