To prove a the given below set is a generating set in my example Consider a set, say $A=\{1,2,3,4,5,6\}$. Let $\mathcal{P}(A)$ be the power set of $A$. Consider 
$$ S=\big\{ \{1,2,5\}, \{2,3,6\}, \{3,4,1\}, \{4,5,2\}, \{5,6,3\}, \{6,1,4\} \big\} $$
We have to prove that the set $S$ generate or we have to check whether set $S$ enumerates all the elements of the set $\mathcal{P}(A)$ under the operation symmetric diference on $S$ that is taking all possible mutual set symmetric difference between the sets taking $2$ sets, $3$ sets, $4$ sets, $5$ sets, $6$ sets at a time in different order as the operation is not associative and taking all their union and along with it original elements of $S$ to be taken union and the null set.
Should give me the power set or the cardinality of this set should be same as the cardinality of the power set $\mathcal{P}(A)$.
We should not use exchaustive method to prove as it may be small. How to prove this in a mathematical proof?
 A: A counterexample to the statement that such an $S$ always generates $\mathcal P(A)$: if we take $n = 6$ and begin with the set $\{1,2,3\}$, then the resulting $S$ will fail to generate $\mathcal P(A)$.
To see that this is the case, it suffices to note that the polynomials $x^2 + x + 1$ and $x^6 - 1$ have a common root (per the discussion below), which means that the matrix $M$ has determinant $0$, which is equal to $0$ modulo $2$.  More generally, the $S$ produced by starting with $\{1,2,3\}$ will fail to generate $\mathcal P(A)$ for any $n$ that is divisible by $3$.

Below are my thoughts on the problem.
The question, as I currently understand it:

Pick an arbitrary integer $n$, and let $A = \{1,\dots,n\}$. Take $S \subset \mathcal P(A)$ to be the orbit of an initial $3$-element set (for instance, $\{1,3,4\}$) under the operation of adding $1$ modulo $n$.  Will the resulting $S$ generate all of $\mathcal P(A)$ under symmetric differences?

Here is an equivalent formulation:

Take an arbitrary vector $v \in \Bbb F_2^n$ that has exactly $3$ non-zero entries.  Let $P$ denote the cyclic permutation matrix given by
  $$
P = \begin{pmatrix}
 0&0&\cdots&0&1\\
 1&0&\cdots&0&0\\
 0&\ddots&\ddots&\vdots&\vdots\\
 \vdots&\ddots&\ddots&0&0\\
 0&\cdots&0&1&0
\end{pmatrix}.
$$
  Do the vectors $\{v,Pv,\dots,P^{n-1}v\}$ span $\Bbb F_2^n$?

If we take the above vectors as the rows of a matrix $M$, then we end up with the matrix described by
$$
m_{ij} = \begin{cases} 1 & \text{set }i\text{ contains element } j\\ 0 & \text{otherwise} \end{cases}.
$$
This gives us a circulant matrix, as exemplified in the comments on the question above.
One approach to answering the question is to compute the determinant of the circulant matrix modulo $2$.  Using the formula given here, we find that this determinant can be expressed as follows.  Let $D = \{d_1,d_2,d_3\} \subset \{1,\dots,n\}$ denote the initial set whose orbit yields $S$, and let $f(x) = \sum_{d \in D} x^{d-1}$.  Then we have
$$
\det M = \prod_{j=0}^{n-1} f(\omega^j)
$$
where $\omega = e^{2 \pi i/n}$.  So, determining the answer to your question is equivalent to determining whether the above product (which must be an integer) is necessarily odd.
It is notable that for any $n$, both $f(\omega^0) = f(1)$ and $f(-1)$ will be odd integers.  Thus, they will not affect the parity of the above product.  With that in mind, it suffices to determine whether the product
$$
Q = \prod_{j:\omega^j \notin \{\pm 1\}} f(\omega^j)
$$
is odd.  By pairing conjugates together in the above product, we have
$$
Q = \prod_{j=1}^{\lfloor (n-1)/2\rfloor} f(\omega^j)\overline{f(\omega^j)} = 
\prod_{j=1}^{\lfloor (n-1)/2\rfloor} |f(\omega^j)|^2
$$
