YES(?,O(n^3))

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

Proof:
 RT Transformation Processor:
  strict:
   b(c(x1)) -> c(b(x1))
   c(b(x1)) -> a(a(a(x1)))
   a(a(a(a(x1)))) -> b(c(x1))
  weak:
   
  Matrix Interpretation Processor:
   dimension: 1
   interpretation:
    [a](x0) = x0 + 6,
    
    [b](x0) = x0,
    
    [c](x0) = x0 + 3
   orientation:
    b(c(x1)) = x1 + 3 >= x1 + 3 = c(b(x1))
    
    c(b(x1)) = x1 + 3 >= x1 + 18 = a(a(a(x1)))
    
    a(a(a(a(x1)))) = x1 + 24 >= x1 + 3 = b(c(x1))
   problem:
    strict:
     b(c(x1)) -> c(b(x1))
     c(b(x1)) -> a(a(a(x1)))
    weak:
     a(a(a(a(x1)))) -> b(c(x1))
   Matrix Interpretation Processor:
    dimension: 1
    interpretation:
     [a](x0) = x0 + 2,
     
     [b](x0) = x0 + 8,
     
     [c](x0) = x0
    orientation:
     b(c(x1)) = x1 + 8 >= x1 + 8 = c(b(x1))
     
     c(b(x1)) = x1 + 8 >= x1 + 6 = a(a(a(x1)))
     
     a(a(a(a(x1)))) = x1 + 8 >= x1 + 8 = b(c(x1))
    problem:
     strict:
      b(c(x1)) -> c(b(x1))
     weak:
      c(b(x1)) -> a(a(a(x1)))
      a(a(a(a(x1)))) -> b(c(x1))
    Matrix Interpretation Processor:
     dimension: 3
     interpretation:
                [1 1 0]     [0]
      [a](x0) = [0 1 1]x0 + [0]
                [0 0 0]     [4],
      
                [1 3 3]     [0]
      [b](x0) = [0 1 1]x0 + [6]
                [0 0 0]     [4],
      
                     [6]
      [c](x0) = x0 + [2]
                     [0]
     orientation:
                 [1 3 3]     [12]    [1 3 3]     [6]           
      b(c(x1)) = [0 1 1]x1 + [8 ] >= [0 1 1]x1 + [8] = c(b(x1))
                 [0 0 0]     [4 ]    [0 0 0]     [4]           
      
                 [1 3 3]     [6]    [1 3 2]     [4]              
      c(b(x1)) = [0 1 1]x1 + [8] >= [0 1 1]x1 + [8] = a(a(a(x1)))
                 [0 0 0]     [4]    [0 0 0]     [4]              
      
                       [1 4 3]     [12]    [1 3 3]     [12]           
      a(a(a(a(x1)))) = [0 1 1]x1 + [12] >= [0 1 1]x1 + [8 ] = b(c(x1))
                       [0 0 0]     [4 ]    [0 0 0]     [4 ]           
     problem:
      strict:
       
      weak:
       b(c(x1)) -> c(b(x1))
       c(b(x1)) -> a(a(a(x1)))
       a(a(a(a(x1)))) -> b(c(x1))
     Qed