YES(?,O(n^3))

Problem:
 a(b(c(x1))) -> c(b(a(x1)))
 C(B(A(x1))) -> A(B(C(x1)))
 b(a(C(x1))) -> C(a(b(x1)))
 c(A(B(x1))) -> B(A(c(x1)))
 A(c(b(x1))) -> b(c(A(x1)))
 B(C(a(x1))) -> a(C(B(x1)))
 a(A(x1)) -> x1
 A(a(x1)) -> x1
 b(B(x1)) -> x1
 B(b(x1)) -> x1
 c(C(x1)) -> x1
 C(c(x1)) -> x1

Proof:
 Matrix Interpretation Processor:
  dimension: 3
  interpretation:
             [1 1 0]     [0]
   [C](x0) = [0 1 0]x0 + [0]
             [0 0 1]     [2],
   
                  [0]
   [B](x0) = x0 + [0]
                  [2],
   
             [1 3 0]     [1]
   [A](x0) = [0 1 0]x0 + [2]
             [0 0 1]     [0],
   
                  [0]
   [a](x0) = x0 + [1]
                  [0],
   
             [1 0 1]     [3]
   [b](x0) = [0 1 0]x0 + [0]
             [0 0 1]     [0],
   
             [1 1 4]     [0]
   [c](x0) = [0 1 0]x0 + [3]
             [0 0 1]     [2]
  orientation:
                 [1 1 5]     [5]    [1 1 5]     [4]              
   a(b(c(x1))) = [0 1 0]x1 + [4] >= [0 1 0]x1 + [4] = c(b(a(x1)))
                 [0 0 1]     [2]    [0 0 1]     [2]              
   
                 [1 4 0]     [3]    [1 4 0]     [1]              
   C(B(A(x1))) = [0 1 0]x1 + [2] >= [0 1 0]x1 + [2] = A(B(C(x1)))
                 [0 0 1]     [4]    [0 0 1]     [4]              
   
                 [1 1 1]     [5]    [1 1 1]     [4]              
   b(a(C(x1))) = [0 1 0]x1 + [1] >= [0 1 0]x1 + [1] = C(a(b(x1)))
                 [0 0 1]     [2]    [0 0 1]     [2]              
   
                 [1 4 4]     [11]    [1 4 4]     [10]              
   c(A(B(x1))) = [0 1 0]x1 + [5 ] >= [0 1 0]x1 + [5 ] = B(A(c(x1)))
                 [0 0 1]     [4 ]    [0 0 1]     [4 ]              
   
                 [1 4 5]     [13]    [1 4 5]     [8]              
   A(c(b(x1))) = [0 1 0]x1 + [5 ] >= [0 1 0]x1 + [5] = b(c(A(x1)))
                 [0 0 1]     [2 ]    [0 0 1]     [2]              
   
                 [1 1 0]     [1]    [1 1 0]     [0]              
   B(C(a(x1))) = [0 1 0]x1 + [1] >= [0 1 0]x1 + [1] = a(C(B(x1)))
                 [0 0 1]     [4]    [0 0 1]     [4]              
   
              [1 3 0]     [1]           
   a(A(x1)) = [0 1 0]x1 + [3] >= x1 = x1
              [0 0 1]     [0]           
   
              [1 3 0]     [4]           
   A(a(x1)) = [0 1 0]x1 + [3] >= x1 = x1
              [0 0 1]     [0]           
   
              [1 0 1]     [5]           
   b(B(x1)) = [0 1 0]x1 + [0] >= x1 = x1
              [0 0 1]     [2]           
   
              [1 0 1]     [3]           
   B(b(x1)) = [0 1 0]x1 + [0] >= x1 = x1
              [0 0 1]     [2]           
   
              [1 2 4]     [8]           
   c(C(x1)) = [0 1 0]x1 + [3] >= x1 = x1
              [0 0 1]     [4]           
   
              [1 2 4]     [3]           
   C(c(x1)) = [0 1 0]x1 + [3] >= x1 = x1
              [0 0 1]     [4]           
  problem:
   
  Qed