YES

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

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