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]     [0]
   [a](x0) = [0 0 0 0]x0 + [0]
             [0 1 1 0]     [1],
   
             [1 0 1 0]     [0]
             [0 0 1 0]     [0]
   [c](x0) = [0 1 0 0]x0 + [0]
             [0 0 1 0]     [1],
   
             [1 0 0 0]     [0]
             [0 0 0 1]     [0]
   [b](x0) = [0 0 0 0]x0 + [0]
             [0 0 0 1]     [1]
  orientation:
                 [1 0 0 1]     [0]    [1 0 0 1]     [0]              
                 [0 0 0 1]     [0]    [0 0 0 1]     [0]              
   c(c(b(x1))) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = a(c(b(x1)))
                 [0 0 0 1]     [1]    [0 0 0 1]     [1]              
   
                    [1 1 2 0]     [1]    [1 1 1 0]     [0]              
                    [0 1 1 0]     [1]    [0 1 0 0]     [1]              
   a(c(b(a(x1)))) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = b(c(c(x1)))
                    [0 1 1 0]     [2]    [0 1 0 0]     [2]              
   
                 [1 1 1 0]     [0]    [1 0 1 0]     [0]                 
                 [0 1 1 0]     [1]    [0 0 0 0]     [0]                 
   b(a(c(x1))) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = a(b(c(a(x1))))
                 [0 1 1 0]     [2]    [0 0 0 0]     [2]                 
   
                 [1 0 1 0]     [0]    [1 0 0 0]     [0]              
                 [0 0 0 0]     [1]    [0 0 0 0]     [0]              
   b(c(a(x1))) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = c(a(b(x1)))
                 [0 0 0 0]     [2]    [0 0 0 0]     [1]              
  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)))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [a](x0) = x0,
    
    [c](x0) = 2x0,
    
    [b](x0) = 2x0
   orientation:
    c(c(b(x1))) = 6x1 >= 4x1 = a(c(b(x1)))
    
    b(a(c(x1))) = 4x1 >= 4x1 = a(b(c(a(x1))))
    
    b(c(a(x1))) = 4x1 >= 4x1 = c(a(b(x1)))
   problem:
    b(a(c(x1))) -> a(b(c(a(x1))))
    b(c(a(x1))) -> c(a(b(x1)))
   String Reversal Processor:
    c(a(b(x1))) -> a(c(b(a(x1))))
    a(c(b(x1))) -> b(a(c(x1)))
    Bounds Processor:
     bound: 3
     enrichment: match
     automaton:
      final states: {6,1}
      transitions:
       c3(48) -> 49*
       b1(26) -> 27*
       b1(11) -> 12*
       b1(23) -> 24*
       a1(25) -> 26*
       a1(10) -> 11*
       a1(22) -> 23*
       a1(28) -> 29*
       c1(27) -> 28*
       c1(9) -> 10*
       c1(21) -> 22*
       b2(37) -> 38*
       b2(44) -> 45*
       a2(46) -> 47*
       a2(36) -> 37*
       a2(43) -> 44*
       f30() -> 2*
       c2(45) -> 46*
       c2(35) -> 36*
       a0(5) -> 1*
       a0(7) -> 8*
       a0(2) -> 3*
       b3(50) -> 51*
       c0(2) -> 7*
       c0(4) -> 5*
       a3(49) -> 50*
       b0(8) -> 6*
       b0(3) -> 4*
       1 -> 7,10
       3 -> 9*
       6 -> 3,9,8
       8 -> 21*
       11 -> 25*
       12 -> 1*
       23 -> 43*
       24 -> 23,11
       26 -> 35*
       29 -> 22*
       38 -> 29*
       44 -> 48*
       47 -> 49,36
       51 -> 47,36
     problem:
      
     Qed