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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a#_A(x1) = ((1,0),(0,0)) x1
          a_A(x1) = ((1,0),(1,0)) x1 + (2,0)
          f_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (0,1)
          b_A() = (2,0)
  
    precedence:
    
      a# = a = f = b
    
    partial status:
  
      pi(a#) = []
      pi(a) = []
      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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a#_A(x1) = ((0,1),(0,0)) x1 + (3,4)
          a_A(x1) = (4,3)
          f_A(x1,x2) = ((0,1),(0,0)) x1 + (1,2)
          b_A() = (7,1)
  
    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.

-- 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a#_A(x1) = ((1,0),(0,0)) x1 + (1,5)
          a_A(x1) = ((0,1),(1,0)) x1 + (4,2)
          f_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (1,1)
          b_A() = (11,1)
  
    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.