YES

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

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