YES

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

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