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

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

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

The next rules are strictly ordered:

  p1

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