YES

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

Proof:
 DP Processor:
  DPs:
   a#(a(x1)) -> b#(x1)
   a#(a(x1)) -> b#(b(x1))
   a#(a(x1)) -> b#(b(b(x1)))
   a#(a(x1)) -> a#(b(b(b(x1))))
   b#(a(x1)) -> b#(c(x1))
   b#(a(x1)) -> b#(b(c(x1)))
   a#(b(b(c(x1)))) -> b#(x1)
   a#(b(b(c(x1)))) -> a#(b(x1))
   a#(b(b(c(x1)))) -> a#(a(b(x1)))
   a#(b(b(c(x1)))) -> a#(a(a(b(x1))))
  TRS:
   a(a(x1)) -> a(b(b(b(x1))))
   b(a(x1)) -> b(b(c(x1)))
   a(b(b(c(x1)))) -> a(a(a(b(x1))))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [b#](x0) = [0 0 1]x0,
    
    [a#](x0) = [3 0 1]x0,
    
              [0 0 0]     [0]
    [c](x0) = [0 1 1]x0 + [1]
              [0 0 0]     [0],
    
              [0 0 2]  
    [b](x0) = [0 0 0]x0
              [0 1 0]  ,
    
              [0 0 0]     [0]
    [a](x0) = [0 0 0]x0 + [0]
              [2 1 1]     [2]
   orientation:
    a#(a(x1)) = [2 1 1]x1 + [2] >= [0 0 1]x1 = b#(x1)
    
    a#(a(x1)) = [2 1 1]x1 + [2] >= [0 1 0]x1 = b#(b(x1))
    
    a#(a(x1)) = [2 1 1]x1 + [2] >= [0] = b#(b(b(x1)))
    
    a#(a(x1)) = [2 1 1]x1 + [2] >= [0] = a#(b(b(b(x1))))
    
    b#(a(x1)) = [2 1 1]x1 + [2] >= [0] = b#(c(x1))
    
    b#(a(x1)) = [2 1 1]x1 + [2] >= [0 1 1]x1 + [1] = b#(b(c(x1)))
    
    a#(b(b(c(x1)))) = [0 6 6]x1 + [6] >= [0 0 1]x1 = b#(x1)
    
    a#(b(b(c(x1)))) = [0 6 6]x1 + [6] >= [0 1 6]x1 = a#(b(x1))
    
    a#(b(b(c(x1)))) = [0 6 6]x1 + [6] >= [0 1 4]x1 + [2] = a#(a(b(x1)))
    
    a#(b(b(c(x1)))) = [0 6 6]x1 + [6] >= [0 1 4]x1 + [4] = a#(a(a(b(x1))))
    
               [0 0 0]     [0]    [0]                 
    a(a(x1)) = [0 0 0]x1 + [0] >= [0] = a(b(b(b(x1))))
               [2 1 1]     [4]    [2]                 
    
               [4 2 2]     [4]    [0 2 2]     [2]              
    b(a(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(b(c(x1)))
               [0 0 0]     [0]    [0 0 0]     [0]              
    
                     [0 0 0]     [0]    [0 0 0]     [0]                 
    a(b(b(c(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(a(a(b(x1))))
                     [0 4 4]     [6]    [0 1 4]     [6]                 
   problem:
    DPs:
     
    TRS:
     a(a(x1)) -> a(b(b(b(x1))))
     b(a(x1)) -> b(b(c(x1)))
     a(b(b(c(x1)))) -> a(a(a(b(x1))))
   Qed