YES

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

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