YES(?,O(n^2))

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

Proof:
 RT Transformation Processor:
  strict:
   b(c(x1)) -> a(x1)
   b(b(x1)) -> a(c(x1))
   a(x1) -> c(b(x1))
   c(c(c(x1))) -> b(x1)
  weak:
   
  Matrix Interpretation Processor:
   dimension: 1
   interpretation:
    [a](x0) = x0 + 8,
    
    [b](x0) = x0,
    
    [c](x0) = x0 + 5
   orientation:
    b(c(x1)) = x1 + 5 >= x1 + 8 = a(x1)
    
    b(b(x1)) = x1 >= x1 + 13 = a(c(x1))
    
    a(x1) = x1 + 8 >= x1 + 5 = c(b(x1))
    
    c(c(c(x1))) = x1 + 15 >= x1 = b(x1)
   problem:
    strict:
     b(c(x1)) -> a(x1)
     b(b(x1)) -> a(c(x1))
    weak:
     a(x1) -> c(b(x1))
     c(c(c(x1))) -> b(x1)
   Matrix Interpretation Processor:
    dimension: 1
    interpretation:
     [a](x0) = x0 + 23,
     
     [b](x0) = x0 + 16,
     
     [c](x0) = x0 + 7
    orientation:
     b(c(x1)) = x1 + 23 >= x1 + 23 = a(x1)
     
     b(b(x1)) = x1 + 32 >= x1 + 30 = a(c(x1))
     
     a(x1) = x1 + 23 >= x1 + 23 = c(b(x1))
     
     c(c(c(x1))) = x1 + 21 >= x1 + 16 = b(x1)
    problem:
     strict:
      b(c(x1)) -> a(x1)
     weak:
      b(b(x1)) -> a(c(x1))
      a(x1) -> c(b(x1))
      c(c(c(x1))) -> b(x1)
    Matrix Interpretation Processor:
     dimension: 2
     interpretation:
                [1 4]     [8]
      [a](x0) = [0 1]x0 + [9],
      
                [1 3]     [2]
      [b](x0) = [0 1]x0 + [6],
      
                [1 1]     [0]
      [c](x0) = [0 1]x0 + [3]
     orientation:
                 [1 4]     [11]    [1 4]     [8]        
      b(c(x1)) = [0 1]x1 + [9 ] >= [0 1]x1 + [9] = a(x1)
      
                 [1 6]     [22]    [1 5]     [20]           
      b(b(x1)) = [0 1]x1 + [12] >= [0 1]x1 + [12] = a(c(x1))
      
              [1 4]     [8]    [1 4]     [8]           
      a(x1) = [0 1]x1 + [9] >= [0 1]x1 + [9] = c(b(x1))
      
                    [1 3]     [9]    [1 3]     [2]        
      c(c(c(x1))) = [0 1]x1 + [9] >= [0 1]x1 + [6] = b(x1)
     problem:
      strict:
       
      weak:
       b(c(x1)) -> a(x1)
       b(b(x1)) -> a(c(x1))
       a(x1) -> c(b(x1))
       c(c(c(x1))) -> b(x1)
     Qed