YES

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

Proof:
 Matrix Interpretation Processor: dim=4
  
  interpretation:
             [1 0 1 0]     [0]
             [0 0 1 0]     [1]
   [c](x0) = [0 0 0 1]x0 + [0]
             [0 0 0 1]     [0],
   
             [1 0 0 0]     [0]
             [0 1 0 0]     [1]
   [b](x0) = [0 0 0 0]x0 + [0]
             [0 1 0 0]     [0],
   
             [1 0 1 0]     [0]
             [0 0 0 1]     [1]
   [a](x0) = [0 0 0 0]x0 + [0]
             [0 0 1 0]     [0]
  orientation:
                 [1 1 0 0]     [0]    [1 1 0 0]     [0]              
                 [0 1 0 0]     [1]    [0 1 0 0]     [1]              
   c(c(b(x1))) = [0 1 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = a(c(b(x1)))
                 [0 1 0 0]     [0]    [0 1 0 0]     [0]              
   
                    [1 0 1 1]     [1]    [1 0 1 1]     [0]              
                    [0 0 0 1]     [2]    [0 0 0 1]     [2]              
   a(c(b(a(x1)))) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = b(c(c(x1)))
                    [0 0 0 1]     [1]    [0 0 0 1]     [1]              
   
                 [1 0 1 1]     [0]    [1 0 1 0]     [0]                 
                 [0 0 0 1]     [2]    [0 0 0 0]     [2]                 
   b(a(c(x1))) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = a(b(c(a(x1))))
                 [0 0 0 1]     [1]    [0 0 0 0]     [0]                 
   
                 [1 0 1 0]     [0]    [1 0 0 0]     [0]              
                 [0 0 0 0]     [2]    [0 0 0 0]     [1]              
   b(c(a(x1))) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = c(a(b(x1)))
                 [0 0 0 0]     [1]    [0 0 0 0]     [0]              
  problem:
   c(c(b(x1))) -> a(c(b(x1)))
   b(a(c(x1))) -> a(b(c(a(x1))))
   b(c(a(x1))) -> c(a(b(x1)))
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {9,5,1}
    transitions:
     c0(3) -> 4*
     c0(6) -> 7*
     c0(10) -> 9*
     c1(13) -> 14*
     b0(2) -> 3*
     b0(7) -> 8*
     a0(3) -> 10*
     a0(4) -> 1*
     a0(2) -> 6*
     a0(8) -> 5*
     f30() -> 2*
     a1(12) -> 13*
     b1(11) -> 12*
     14 -> 8*
     2 -> 11*
     5 -> 12,3
     9 -> 12,3
   problem:
    
   Qed