YES

We show the termination of the TRS R:

  a(a(x)) -> b(b(x))
  b(b(a(x))) -> a(b(b(x)))

-- SCC decomposition.

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

p1: a#(a(x)) -> b#(b(x))
p2: a#(a(x)) -> b#(x)
p3: b#(b(a(x))) -> a#(b(b(x)))
p4: b#(b(a(x))) -> b#(b(x))
p5: b#(b(a(x))) -> b#(x)

and R consists of:

r1: a(a(x)) -> b(b(x))
r2: b(b(a(x))) -> a(b(b(x)))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: a#(a(x)) -> b#(b(x))
p2: b#(b(a(x))) -> b#(x)
p3: b#(b(a(x))) -> b#(b(x))
p4: b#(b(a(x))) -> a#(b(b(x)))
p5: a#(a(x)) -> b#(x)

and R consists of:

r1: a(a(x)) -> b(b(x))
r2: b(b(a(x))) -> a(b(b(x)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a#_A(x1) = ((1,0),(1,1)) x1
          a_A(x1) = ((1,0),(1,0)) x1 + (11,9)
          b#_A(x1) = ((1,0),(0,0)) x1 + (1,8)
          b_A(x1) = ((1,0),(1,0)) x1 + (2,1)
  
    precedence:
    
      a# = a = b# = b
    
    partial status:
  
      pi(a#) = [1]
      pi(a) = []
      pi(b#) = []
      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: b#(b(a(x))) -> b#(x)
p2: b#(b(a(x))) -> b#(b(x))
p3: b#(b(a(x))) -> a#(b(b(x)))
p4: a#(a(x)) -> b#(x)

and R consists of:

r1: a(a(x)) -> b(b(x))
r2: b(b(a(x))) -> a(b(b(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: b#(b(a(x))) -> b#(x)
p2: b#(b(a(x))) -> a#(b(b(x)))
p3: a#(a(x)) -> b#(x)
p4: b#(b(a(x))) -> b#(b(x))

and R consists of:

r1: a(a(x)) -> b(b(x))
r2: b(b(a(x))) -> a(b(b(x)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          b#_A(x1) = ((0,0),(1,0)) x1 + (1,2)
          b_A(x1) = ((1,0),(0,0)) x1 + (2,7)
          a_A(x1) = ((1,0),(0,0)) x1 + (2,7)
          a#_A(x1) = ((0,0),(1,0)) x1 + (1,1)
  
    precedence:
    
      b# = b = a > a#
    
    partial status:
  
      pi(b#) = []
      pi(b) = []
      pi(a) = []
      pi(a#) = []

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: b#(b(a(x))) -> b#(x)
p2: b#(b(a(x))) -> a#(b(b(x)))
p3: b#(b(a(x))) -> b#(b(x))

and R consists of:

r1: a(a(x)) -> b(b(x))
r2: b(b(a(x))) -> a(b(b(x)))

The estimated dependency graph contains the following SCCs:

  {p1, p3}


-- Reduction pair.

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

p1: b#(b(a(x))) -> b#(x)
p2: b#(b(a(x))) -> b#(b(x))

and R consists of:

r1: a(a(x)) -> b(b(x))
r2: b(b(a(x))) -> a(b(b(x)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          b#_A(x1) = ((0,0),(1,0)) x1 + (6,3)
          b_A(x1) = ((1,0),(0,0)) x1 + (2,2)
          a_A(x1) = ((1,0),(0,0)) x1 + (3,1)
  
    precedence:
    
      b# = b = a
    
    partial status:
  
      pi(b#) = []
      pi(b) = []
      pi(a) = []

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: b#(b(a(x))) -> b#(b(x))

and R consists of:

r1: a(a(x)) -> b(b(x))
r2: b(b(a(x))) -> a(b(b(x)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: b#(b(a(x))) -> b#(b(x))

and R consists of:

r1: a(a(x)) -> b(b(x))
r2: b(b(a(x))) -> a(b(b(x)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          b#_A(x1) = x1 + (1,1)
          b_A(x1) = x1 + (1,1)
          a_A(x1) = x1 + (3,3)
  
    precedence:
    
      b# = b > a
    
    partial status:
  
      pi(b#) = []
      pi(b) = [1]
      pi(a) = [1]

The next rules are strictly ordered:

  p1

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