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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a#_A(x1) = ((1,1),(0,0)) x1
        a_A(x1) = ((1,0),(1,1)) x1 + (8,1)
        f_A(x1,x2) = x1 + x2 + (1,1)
        f#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (1,0)
        b_A(x1) = x1 + (0,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a#_A(x1) = (1,1)
        a_A(x1) = (2,4)
        f_A(x1,x2) = ((0,0),(1,0)) x1 + (1,1)
        f#_A(x1,x2) = (0,0)
        b_A(x1) = (2,4)
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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