YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 3
  interpretation:
             [0  -& 0 ]  
   [b](x0) = [1  0  1 ]x0
             [0  -& 0 ]  ,
   
             [0  1  -&]  
   [c](x0) = [0  2  -&]x0
             [0  1  -&]  ,
   
             [1  0  -&]  
   [a](x0) = [0  -& 0 ]x0
             [1  2  -&]  
  orientation:
              [2  1  0 ]      [0  1  -&]                
   a(a(x1)) = [1  2  -&]x1 >= [1  2  -&]x1 = b(b(c(x1)))
              [2  1  2 ]      [0  1  -&]                
   
              [1  2  -&]      [0  1  -&]          
   b(a(x1)) = [2  3  0 ]x1 >= [0  2  -&]x1 = c(x1)
              [1  2  -&]      [0  1  -&]          
   
              [2 1 2]      [2  1  0 ]             
   c(b(x1)) = [3 2 3]x1 >= [1  2  -&]x1 = a(a(x1))
              [2 1 2]      [2  1  2 ]             
  problem:
   a(a(x1)) -> b(b(c(x1)))
   c(b(x1)) -> a(a(x1))
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {5,1}
    transitions:
     f30() -> 2*
     b0(4) -> 1*
     b0(3) -> 4*
     c0(2) -> 3*
     a0(2) -> 6*
     a0(6) -> 5*
     b1(12) -> 13*
     b1(13) -> 14*
     c1(11) -> 12*
     1 -> 6*
     2 -> 11*
     5 -> 12,3
     14 -> 5*
   problem:
    
   Qed