YES

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

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