YES

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

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
             [1 1]  
   [a](x0) = [0 1]x0,
   
             [1 1]     [0]
   [b](x0) = [0 3]x0 + [1],
   
             [1 0]  
   [c](x0) = [0 0]x0
  orientation:
              [1 4]     [1]    [1 4]     [0]                 
   a(b(x1)) = [0 3]x1 + [1] >= [0 3]x1 + [1] = b(a(a(a(x1))))
   
              [1 2]     [0]    [1 2]             
   b(a(x1)) = [0 3]x1 + [1] >= [0 1]x1 = a(a(x1))
   
              [1 2]      [1 1]                
   a(a(x1)) = [0 1]x1 >= [0 0]x1 = a(c(b(x1)))
  problem:
   b(a(x1)) -> a(a(x1))
   a(a(x1)) -> a(c(b(x1)))
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {4,1}
    transitions:
     c0(5) -> 6*
     a1(9) -> 10*
     b0(2) -> 5*
     a0(3) -> 1*
     a0(6) -> 4*
     a0(2) -> 3*
     f30() -> 2*
     c1(8) -> 9*
     b1(15) -> 16*
     b1(7) -> 8*
     16 -> 8*
     4 -> 3*
     2 -> 7*
     6 -> 15*
     1 -> 8,5
     10 -> 1*
   problem:
    
   Qed