YES

Problem:
 a(b(b(x1))) -> b(a(a(x1)))
 a(a(b(x1))) -> b(b(a(x1)))

Proof:
 DP Processor:
  DPs:
   a#(b(b(x1))) -> a#(x1)
   a#(b(b(x1))) -> a#(a(x1))
   a#(a(b(x1))) -> a#(x1)
  TRS:
   a(b(b(x1))) -> b(a(a(x1)))
   a(a(b(x1))) -> b(b(a(x1)))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [a#](x0) = 2x0 + 0,
    
    [a](x0) = 1x0 + 7,
    
    [b](x0) = 1x0 + 12
   orientation:
    a#(b(b(x1))) = 4x1 + 15 >= 2x1 + 0 = a#(x1)
    
    a#(b(b(x1))) = 4x1 + 15 >= 3x1 + 9 = a#(a(x1))
    
    a#(a(b(x1))) = 4x1 + 15 >= 2x1 + 0 = a#(x1)
    
    a(b(b(x1))) = 3x1 + 14 >= 3x1 + 12 = b(a(a(x1)))
    
    a(a(b(x1))) = 3x1 + 14 >= 3x1 + 13 = b(b(a(x1)))
   problem:
    DPs:
     
    TRS:
     a(b(b(x1))) -> b(a(a(x1)))
     a(a(b(x1))) -> b(b(a(x1)))
   Qed