YES

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

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