YES

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

Proof:
 String Reversal Processor:
  a(a(x1)) -> b(x1)
  a(b(x1)) -> b(a(x1))
  c(b(b(x1))) -> a(c(x1))
  b(b(x1)) -> a(a(a(x1)))
  a(c(x1)) -> c(a(b(x1)))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
              [1 0]     [1]
    [c](x0) = [0 3]x0 + [2],
    
              [1 3]     [3]
    [b](x0) = [0 1]x0 + [0],
    
              [1 2]     [2]
    [a](x0) = [0 1]x0 + [0]
   orientation:
               [1 4]     [4]    [1 3]     [3]        
    a(a(x1)) = [0 1]x1 + [0] >= [0 1]x1 + [0] = b(x1)
    
               [1 5]     [5]    [1 5]     [5]           
    a(b(x1)) = [0 1]x1 + [0] >= [0 1]x1 + [0] = b(a(x1))
    
                  [1 6]     [7]    [1 6]     [7]           
    c(b(b(x1))) = [0 3]x1 + [2] >= [0 3]x1 + [2] = a(c(x1))
    
               [1 6]     [6]    [1 6]     [6]              
    b(b(x1)) = [0 1]x1 + [0] >= [0 1]x1 + [0] = a(a(a(x1)))
    
               [1 6]     [7]    [1 5]     [6]              
    a(c(x1)) = [0 3]x1 + [2] >= [0 3]x1 + [2] = c(a(b(x1)))
   problem:
    a(b(x1)) -> b(a(x1))
    c(b(b(x1))) -> a(c(x1))
    b(b(x1)) -> a(a(a(x1)))
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0 0]  
     [c](x0) = [2 1]x0,
     
               [2 0]  
     [b](x0) = [0 0]x0,
     
               [0  -&]  
     [a](x0) = [1  0 ]x0
    orientation:
                [2 0]      [2 0]             
     a(b(x1)) = [3 1]x1 >= [1 0]x1 = b(a(x1))
     
                   [4 2]      [0 0]             
     c(b(b(x1))) = [6 4]x1 >= [2 1]x1 = a(c(x1))
     
                [4 2]      [0  -&]                
     b(b(x1)) = [2 0]x1 >= [1  0 ]x1 = a(a(a(x1)))
    problem:
     a(b(x1)) -> b(a(x1))
     b(b(x1)) -> a(a(a(x1)))
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
                [1 1]  
      [b](x0) = [0 0]x0,
      
                [0 0]  
      [a](x0) = [0 0]x0
     orientation:
                 [1 1]      [1 1]             
      a(b(x1)) = [1 1]x1 >= [0 0]x1 = b(a(x1))
      
                 [2 2]      [0 0]                
      b(b(x1)) = [1 1]x1 >= [0 0]x1 = a(a(a(x1)))
     problem:
      a(b(x1)) -> b(a(x1))
     KBO Processor:
      weight function:
       w0 = 1
       w(b) = w(a) = 1
      precedence:
       a > b
      problem:
       
      Qed