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?

Using Boolean algebra you consider the Boolean variables


*

*$a,b,c$ with $a=1$ ($1$ stands for $True$) iff $x \in A$. Similarly, with $b$ and $c$.


Now, write for each set the corresponding Boolean expressions


*

*$x \in S_1 = (cb')a' + b(b(a+c)')'(abc)'$

*$x \in S_2 = ab(abc)'+ca'$
Now, the only thing you need to show is that these two expressions are logically equivalent. To do so, you can simply transform them into a normal form (for example here I use disjunctive normal form (DNF)).
If the DNF's are equal, the statements are equivalent:
\begin{eqnarray*} (cb')a' + b(b(a+c)')'(abc)'
& = & a'b'c + b(b'+(a+c))(abc)' \\
& \stackrel{uu'=0, 0+v=v}{=} & a'b'c + (ab+bc)(a'+b'+c') \\
& \stackrel{uu'=0, 0+v=v}{=} & \boxed{a'b'c + abc' + a'bc}
\end{eqnarray*}
\begin{eqnarray*} ab(abc)'+ca'
& = & ab(a'+b'+c') + c(b+b')a' \\
& \stackrel{uu'=0, 0+v=v}{=} & \boxed{abc' + a'bc + a'b'c}
\end{eqnarray*}
They are equal. So, the sets are equal.
