YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: a#(a(f(b(),a(x)))) -> f#(a(a(a(x))),b())
p2: a#(a(f(b(),a(x)))) -> a#(a(a(x)))
p3: a#(a(f(b(),a(x)))) -> a#(a(x))
p4: a#(a(x)) -> f#(b(),a(f(a(x),b())))
p5: a#(a(x)) -> a#(f(a(x),b()))
p6: a#(a(x)) -> f#(a(x),b())
p7: f#(a(x),b()) -> f#(b(),a(x))

and R consists of:

r1: a(a(f(b(),a(x)))) -> f(a(a(a(x))),b())
r2: a(a(x)) -> f(b(),a(f(a(x),b())))
r3: f(a(x),b()) -> f(b(),a(x))

The estimated dependency graph contains the following SCCs:

  {p2, p3, p5}


-- Reduction pair.

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

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

and R consists of:

r1: a(a(f(b(),a(x)))) -> f(a(a(a(x))),b())
r2: a(a(x)) -> f(b(),a(f(a(x),b())))
r3: f(a(x),b()) -> f(b(),a(x))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1) = x1 + 8
          a_A(x1) = x1 + 5
          f_A(x1,x2) = max{2, x1, x2}
          b_A = 4
    
      precedence:
      
        a# = a = f = b
      
      partial status:
    
        pi(a#) = []
        pi(a) = []
        pi(f) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1) = 63
          a_A(x1) = x1 + 34
          f_A(x1,x2) = 28
          b_A = 38
    
      precedence:
      
        a# > a = f = b
      
      partial status:
    
        pi(a#) = []
        pi(a) = [1]
        pi(f) = []
        pi(b) = []
    

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

and R consists of:

r1: a(a(f(b(),a(x)))) -> f(a(a(a(x))),b())
r2: a(a(x)) -> f(b(),a(f(a(x),b())))
r3: f(a(x),b()) -> f(b(),a(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#(a(x)) -> a#(f(a(x),b()))
p2: a#(a(f(b(),a(x)))) -> a#(a(a(x)))

and R consists of:

r1: a(a(f(b(),a(x)))) -> f(a(a(a(x))),b())
r2: a(a(x)) -> f(b(),a(f(a(x),b())))
r3: f(a(x),b()) -> f(b(),a(x))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1) = x1 + 5
          a_A(x1) = 0
          f_A(x1,x2) = max{0, x1 - 6, x2 - 1}
          b_A = 0
    
      precedence:
      
        a# = a = f = b
      
      partial status:
    
        pi(a#) = []
        pi(a) = []
        pi(f) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1) = max{31, x1 - 1}
          a_A(x1) = 33
          f_A(x1,x2) = 25
          b_A = 36
    
      precedence:
      
        b > a > a# > f
      
      partial status:
    
        pi(a#) = []
        pi(a) = []
        pi(f) = []
        pi(b) = []
    

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

and R consists of:

r1: a(a(f(b(),a(x)))) -> f(a(a(a(x))),b())
r2: a(a(x)) -> f(b(),a(f(a(x),b())))
r3: f(a(x),b()) -> f(b(),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: a#(a(f(b(),a(x)))) -> a#(a(a(x)))

and R consists of:

r1: a(a(f(b(),a(x)))) -> f(a(a(a(x))),b())
r2: a(a(x)) -> f(b(),a(f(a(x),b())))
r3: f(a(x),b()) -> f(b(),a(x))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            a#_A(x1) = ((1,0),(0,0)) x1 + (1,4)
            a_A(x1) = ((0,1),(1,0)) x1 + (5,1)
            f_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (3,0)
            b_A() = (2,2)
    
      precedence:
      
        a# = f > a = b
      
      partial status:
    
        pi(a#) = []
        pi(a) = []
        pi(f) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            a#_A(x1) = (1,1)
            a_A(x1) = (2,2)
            f_A(x1,x2) = (4,4)
            b_A() = (3,3)
    
      precedence:
      
        a# = a = f = b
      
      partial status:
    
        pi(a#) = []
        pi(a) = []
        pi(f) = []
        pi(b) = []
    

The next rules are strictly ordered:

  p1

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