YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [a](x0) = x0,
   
   [b](x0) = x0 + 4,
   
   [c](x0) = x0
  orientation:
   a(b(x1)) = x1 + 4 >= x1 + 4 = b(a(a(x1)))
   
   b(x1) = x1 + 4 >= 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)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 1 0]  
    [a](x0) = [0 0 1]x0
              [0 1 0]  ,
    
              [1 0 0]     [0]
    [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]     [1]    [1 1 1]     [0]              
    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 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:
    a(a(x1)) -> a(c(a(x1)))
   Bounds Processor:
    bound: 2
    enrichment: match
    automaton:
     final states: {3}
     transitions:
      c1(9) -> 10*
      c1(12) -> 13*
      c2(21) -> 22*
      a2(20) -> 21*
      a2(22) -> 23*
      a0(3) -> 3*
      a1(10) -> 11*
      a1(8) -> 9*
      c0(3) -> 3*
      11 -> 9,12,3
      13 -> 10*
      3 -> 8*
      10 -> 20*
      23 -> 9*
    problem:
     
    Qed