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:
 String Reversal Processor:
  b(a(a(x1))) -> a(c(b(a(b(x1)))))
  a(b(x1)) -> b(b(a(x1)))
  a(c(b(x1))) -> b(a(c(x1)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 1]     [0]
    [a](x0) = [0 0 0]x0 + [0]
              [1 1 1]     [1],
    
                
    [b](x0) = x0
                ,
    
              [1 0 0]     [0]
    [c](x0) = [0 0 0]x0 + [1]
              [1 0 0]     [0]
   orientation:
                  [2 1 2]     [1]    [2 0 2]     [0]                    
    b(a(a(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(c(b(a(b(x1)))))
                  [2 1 2]     [2]    [2 0 2]     [2]                    
    
               [1 0 1]     [0]    [1 0 1]     [0]              
    a(b(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(b(a(x1)))
               [1 1 1]     [1]    [1 1 1]     [1]              
    
                  [2 0 0]     [0]    [2 0 0]     [0]              
    a(c(b(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(a(c(x1)))
                  [2 0 0]     [2]    [2 0 0]     [2]              
   problem:
    a(b(x1)) -> b(b(a(x1)))
    a(c(b(x1))) -> b(a(c(x1)))
   String Reversal Processor:
    b(a(x1)) -> a(b(b(x1)))
    b(c(a(x1))) -> c(a(b(x1)))
    KBO Processor:
     weight function:
      w0 = 1
      w(c) = w(a) = 1
      w(b) = 0
     precedence:
      b > a > c
     problem:
      
     Qed