YES

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

Proof:
 String Reversal Processor:
  b(a(x1)) -> a(b(b(x1)))
  c(b(x1)) -> b(b(c(x1)))
  a(c(x1)) -> c(c(a(x1)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]  
    [c](x0) = [0 0 0]x0
              [0 0 0]  ,
    
              [1 0 0]     [1]
    [a](x0) = [0 1 1]x0 + [1]
              [0 1 1]     [1],
    
              [1 1 0]  
    [b](x0) = [0 0 1]x0
              [0 1 0]  
   orientation:
               [1 1 1]     [2]    [1 1 1]     [1]              
    b(a(x1)) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = a(b(b(x1)))
               [0 1 1]     [1]    [0 1 1]     [1]              
    
               [1 1 0]      [1 0 0]                
    c(b(x1)) = [0 0 0]x1 >= [0 0 0]x1 = b(b(c(x1)))
               [0 0 0]      [0 0 0]                
    
               [1 0 0]     [1]    [1 0 0]     [1]              
    a(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [0] = c(c(a(x1)))
               [0 0 0]     [1]    [0 0 0]     [0]              
   problem:
    c(b(x1)) -> b(b(c(x1)))
    a(c(x1)) -> c(c(a(x1)))
   Bounds Processor:
    bound: 0
    enrichment: match
    automaton:
     final states: {5,1}
     transitions:
      f30() -> 2*
      b0(4) -> 1*
      b0(3) -> 4*
      c0(7) -> 5*
      c0(2) -> 3*
      c0(6) -> 7*
      a0(2) -> 6*
      1 -> 3*
      5 -> 6*
    problem:
     
    Qed