YES

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

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