YES

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

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