YES

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

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