YES(?,O(n^3))

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

Proof:
 RT Transformation Processor:
  strict:
   a(b(x)) -> b(a(x))
   a(c(x)) -> x
  weak:
   
  Matrix Interpretation Processor:
   dimension: 1
   interpretation:
    [c](x0) = x0,
    
    [a](x0) = x0 + 1,
    
    [b](x0) = x0
   orientation:
    a(b(x)) = x + 1 >= x + 1 = b(a(x))
    
    a(c(x)) = x + 1 >= x = x
   problem:
    strict:
     a(b(x)) -> b(a(x))
    weak:
     a(c(x)) -> x
   Matrix Interpretation Processor:
    dimension: 3
    interpretation:
               [1 0 7]  
     [c](x0) = [0 1 2]x0
               [0 0 1]  ,
     
               [1 1 0]     [0]
     [a](x0) = [0 1 4]x0 + [2]
               [0 0 1]     [0],
     
                    [0]
     [b](x0) = x0 + [4]
                    [2]
    orientation:
               [1 1 0]    [4 ]    [1 1 0]    [0]          
     a(b(x)) = [0 1 4]x + [14] >= [0 1 4]x + [6] = b(a(x))
               [0 0 1]    [2 ]    [0 0 1]    [2]          
     
               [1 1 9]    [0]         
     a(c(x)) = [0 1 6]x + [2] >= x = x
               [0 0 1]    [0]         
    problem:
     strict:
      
     weak:
      a(b(x)) -> b(a(x))
      a(c(x)) -> x
    Qed