YES

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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 0]  
   [b](x0) = [0 1 0]x0
             [0 1 0]  ,
   
             [1 0 1]     [0]
   [a](x0) = [0 1 0]x0 + [1]
             [0 0 0]     [0],
   
             [1 0 0]  
   [c](x0) = [0 0 0]x0
             [0 0 1]  
  orientation:
                    [1 1 1]     [1]    [1 1 0]     [0]                 
   a(b(b(a(x1)))) = [0 1 0]x1 + [2] >= [0 0 0]x1 + [1] = a(c(a(b(x1))))
                    [0 0 0]     [0]    [0 0 0]     [0]                 
   
              [1 0 1]     [0]    [1 0 1]                
   a(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 = c(c(a(x1)))
              [0 0 0]     [0]    [0 0 0]                
   
                 [1 0 0]      [1 0 0]                
   c(c(c(x1))) = [0 0 0]x1 >= [0 0 0]x1 = b(c(b(x1)))
                 [0 0 1]      [0 0 0]                
  problem:
   a(c(x1)) -> c(c(a(x1)))
   c(c(c(x1))) -> b(c(b(x1)))
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {5,1}
    transitions:
     a0(2) -> 3*
     b1(16) -> 17*
     b1(12) -> 13*
     b1(14) -> 15*
     b1(18) -> 19*
     b0(2) -> 6*
     b0(7) -> 5*
     c0(3) -> 4*
     c0(4) -> 1*
     c0(6) -> 7*
     f30() -> 2*
     c1(17) -> 18*
     c1(13) -> 14*
     19 -> 4*
     4 -> 12*
     1 -> 3*
     3 -> 16*
     15 -> 1*
   problem:
    
   Qed