YES

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

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