YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: a#(a(f(x,y))) -> f#(a(b(a(b(a(x))))),a(b(a(b(a(y))))))
p2: a#(a(f(x,y))) -> a#(b(a(b(a(x)))))
p3: a#(a(f(x,y))) -> a#(b(a(x)))
p4: a#(a(f(x,y))) -> a#(x)
p5: a#(a(f(x,y))) -> a#(b(a(b(a(y)))))
p6: a#(a(f(x,y))) -> a#(b(a(y)))
p7: a#(a(f(x,y))) -> a#(y)
p8: f#(a(x),a(y)) -> a#(f(x,y))
p9: f#(a(x),a(y)) -> f#(x,y)
p10: f#(b(x),b(y)) -> f#(x,y)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p4, p7, p8, p9, p10}


-- Reduction pair.

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

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

and R consists of:

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

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: lexicographic order
          interpretations:
            a#_A(x1) = ((0,0),(1,0)) x1 + (4,4)
            a_A(x1) = ((1,0),(1,1)) x1 + (0,4)
            f_A(x1,x2) = x1 + x2 + (3,5)
            f#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (4,7)
            b_A(x1) = ((1,0),(0,0)) x1 + (0,2)
    
      precedence:
      
        a# = a = f# > b > f
      
      partial status:
    
        pi(a#) = []
        pi(a) = []
        pi(f) = [2]
        pi(f#) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = (3,2)
            a_A(x1) = (2,1)
            f_A(x1,x2) = (1,0)
            f#_A(x1,x2) = (3,2)
            b_A(x1) = (2,1)
    
      precedence:
      
        a > a# = f# > b > f
      
      partial status:
    
        pi(a#) = []
        pi(a) = []
        pi(f) = []
        pi(f#) = []
        pi(b) = []
    

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

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(f(x,y))) -> f#(a(b(a(b(a(x))))),a(b(a(b(a(y))))))
p2: f#(a(x),a(y)) -> a#(f(x,y))
p3: a#(a(f(x,y))) -> a#(x)
p4: f#(a(x),a(y)) -> f#(x,y)
p5: f#(b(x),b(y)) -> f#(x,y)

and R consists of:

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

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

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: a#(a(f(x,y))) -> f#(a(b(a(b(a(x))))),a(b(a(b(a(y))))))
p2: f#(a(x),a(y)) -> a#(f(x,y))
p3: f#(a(x),a(y)) -> f#(x,y)
p4: f#(b(x),b(y)) -> f#(x,y)

and R consists of:

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

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#(a(f(x,y))) -> f#(a(b(a(b(a(x))))),a(b(a(b(a(y))))))
p2: f#(a(x),a(y)) -> f#(x,y)
p3: f#(b(x),b(y)) -> f#(x,y)
p4: f#(a(x),a(y)) -> a#(f(x,y))

and R consists of:

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

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) = ((1,1),(1,1)) x1 + (1,1)
            a_A(x1) = ((0,0),(1,1)) x1 + (8,0)
            f_A(x1,x2) = x2
            f#_A(x1,x2) = ((0,1),(0,1)) x2 + (1,2)
            b_A(x1) = ((0,0),(0,1)) x1 + (2,1)
    
      precedence:
      
        a# = a = f = f# > b
      
      partial status:
    
        pi(a#) = []
        pi(a) = []
        pi(f) = []
        pi(f#) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            a#_A(x1) = (0,1)
            a_A(x1) = (0,1)
            f_A(x1,x2) = (0,1)
            f#_A(x1,x2) = (0,1)
            b_A(x1) = (0,1)
    
      precedence:
      
        a# = f# > a = f > b
      
      partial status:
    
        pi(a#) = []
        pi(a) = []
        pi(f) = []
        pi(f#) = []
        pi(b) = []
    

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

and R consists of:

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

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

and R consists of:

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

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

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

and R consists of:

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

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

and R consists of:

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

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: lexicographic order
          interpretations:
            a#_A(x1) = ((0,0),(1,0)) x1 + (2,1)
            a_A(x1) = ((1,0),(0,0)) x1 + (4,9)
            f_A(x1,x2) = x2 + (3,5)
            f#_A(x1,x2) = ((0,0),(1,0)) x2 + (2,2)
            b_A(x1) = (1,10)
    
      precedence:
      
        f# > f = b > a# = a
      
      partial status:
    
        pi(a#) = []
        pi(a) = []
        pi(f) = []
        pi(f#) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a#_A(x1) = (2,2)
            a_A(x1) = (1,0)
            f_A(x1,x2) = (4,2)
            f#_A(x1,x2) = (3,1)
            b_A(x1) = (4,0)
    
      precedence:
      
        b > a# = a = f = f#
      
      partial status:
    
        pi(a#) = []
        pi(a) = []
        pi(f) = []
        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: f#(a(x),a(y)) -> a#(f(x,y))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  (no SCCs)