How to prove using only set identities? I have given three sets A, B and C and I need to prove that following to statements are equivalent:

S1 =  (( − ) − ) ∪ ( − ( − ( ∪ ))) − ( ∩ ( ∩ ))
S2 = ( ∩ ) − ( ∩ ( ∩ )) ∪ ( − )

prove that S1 is equivalent to S2 without using venn diagram.
How can I prove this?

$$S_1=(((C\setminus B)\setminus A)\cup (B\setminus (B\setminus(A\cup C)) \setminus A\cap B\cap C$$
$$S_2=(A\cap B \setminus (A\cap C\cap B)) \cup (C\setminus A)$$
Because $(A\cap C\cap B)) \cap C\setminus A=\emptyset$, we can remove $(A\cap C\cap B))$ from both $S_1$ and $S_2$.
$$S'_1=((C\setminus B)\setminus A)\cup (B\setminus (B\setminus(A\cup C)$$
$$S'_2=(A\cap B)\cup (C\setminus A)$$
Using $X\setminus Y=X\cap Y^c$ we get:
$$S'_1=(C\cap (A\cup B)^c) \cup (B\cap (A\cup C))$$
$$S'_2=(A\cap B)\cup (C\cup A^c)$$
Using De Morgan and distributive laws gives:
$$S'_1=(C\cap A^c\cap B^c) \cup (B\cap (A\cup C))$$
$$S'_2=(A\cup C)\cap (A\cup A^c)\cap (B\cup C) \cap (B\cup A^c)$$
and then:
$$S'_1=(C\cup B)\cap (A^c\cup B)\cap (B^c\cup B)\cap(C\cup (A\cup C))\cap (A^c\cup (A\cup C))\cap (B^c\cup (A\cup C))$$
$$S'_2=(A\cup C)\cap (A\cup A^c)\cap (B\cup C) \cap (B\cup A^c)$$
Removing $\omega$ and like-terms:
$$S''_1=(C\cup (A\cup C))\cap (A^c\cup (A\cup C))\cap (B^c\cup (A\cup C))$$
$$S''_2=A\cup C$$
$S''_1$ further simplifies:
$$S''_1=(A\cup C)\cap (A^c\cup (A\cup C))\cap (B^c\cup (A\cup C))$$
and because $A\subset B \implies A\cap B=A$:
$$S''_1=A\cup C$$
As $S''_1=S''_2$ is true, so is $S_1=S_2$.
