YES

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

Proof:
 DP Processor:
  DPs:
   a#(a(b(x1))) -> a#(x1)
   a#(a(b(x1))) -> b#(c(a(x1)))
   a#(a(b(x1))) -> a#(b(c(a(x1))))
   a#(a(b(x1))) -> b#(a(b(c(a(x1)))))
   b#(a(x1)) -> b#(x1)
   b#(a(x1)) -> b#(b(x1))
   b#(a(x1)) -> a#(b(b(x1)))
   b#(c(a(x1))) -> b#(x1)
   b#(c(a(x1))) -> a#(b(x1))
  TRS:
   a(a(b(x1))) -> b(a(b(c(a(x1)))))
   b(a(x1)) -> a(b(b(x1)))
   b(c(a(x1))) -> c(a(b(x1)))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
    [b#](x0) = [1 0]x0 + [2],
    
    [a#](x0) = [0 2]x0 + [1],
    
              [1 0]  
    [c](x0) = [0 0]x0,
    
              [1 2]     [1]
    [a](x0) = [2 1]x0 + [2],
    
                
    [b](x0) = x0
   orientation:
    a#(a(b(x1))) = [4 2]x1 + [5] >= [0 2]x1 + [1] = a#(x1)
    
    a#(a(b(x1))) = [4 2]x1 + [5] >= [1 2]x1 + [3] = b#(c(a(x1)))
    
    a#(a(b(x1))) = [4 2]x1 + [5] >= [1] = a#(b(c(a(x1))))
    
    a#(a(b(x1))) = [4 2]x1 + [5] >= [1 2]x1 + [4] = b#(a(b(c(a(x1)))))
    
    b#(a(x1)) = [1 2]x1 + [3] >= [1 0]x1 + [2] = b#(x1)
    
    b#(a(x1)) = [1 2]x1 + [3] >= [1 0]x1 + [2] = b#(b(x1))
    
    b#(a(x1)) = [1 2]x1 + [3] >= [0 2]x1 + [1] = a#(b(b(x1)))
    
    b#(c(a(x1))) = [1 2]x1 + [3] >= [1 0]x1 + [2] = b#(x1)
    
    b#(c(a(x1))) = [1 2]x1 + [3] >= [0 2]x1 + [1] = a#(b(x1))
    
                  [5 4]     [6]    [1 2]     [2]                    
    a(a(b(x1))) = [4 5]x1 + [6] >= [2 4]x1 + [4] = b(a(b(c(a(x1)))))
    
               [1 2]     [1]    [1 2]     [1]              
    b(a(x1)) = [2 1]x1 + [2] >= [2 1]x1 + [2] = a(b(b(x1)))
    
                  [1 2]     [1]    [1 2]     [1]              
    b(c(a(x1))) = [0 0]x1 + [0] >= [0 0]x1 + [0] = c(a(b(x1)))
   problem:
    DPs:
     
    TRS:
     a(a(b(x1))) -> b(a(b(c(a(x1)))))
     b(a(x1)) -> a(b(b(x1)))
     b(c(a(x1))) -> c(a(b(x1)))
   Qed