Let A = {a, b, c, e, f} B = {c, d, e, g} and C = {b, c, f, g} be subsets of the set U = {a, b, c, d, e, f, g, h}.
(i) $\mathbf{A} \cap \mathbf{B}$
(ii) $A \cup(B \cap C)$
(iii) $\mathbf{A}-\mathbf{B}$
(iv) $\mathbf{B}-\mathbf{A}$
(v) $A-(B \cap C)$
(vi) $(B-C) \cup(C-B)$
(i) $A^{\cap} B$ will contain the common elements of $A$ and $B$
$\mathrm{A}^{\cap} \mathrm{B}=\{\mathrm{c}, \mathrm{e}\}$
(ii) $A U\left(B^{\cap} C\right)$
$B^{\cap} C=\{c, d, g\}$
$\mathrm{AU}\left(\mathrm{B}^{\cap} \mathrm{C}\right)=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f}, \mathrm{g}\}$
(iii) A - B implies the set of all elements in A that are not in B
$A-B=\{a, b, f\}$
(iv) $B$ - $A$ implies the set of all elements in $B$ that are not in $A$
$\mathrm{B}-\mathrm{A}=\{\mathrm{d}, \mathrm{g}\}$
(v) $A-\left(B^{\cap} C\right)$ denotes elements of $A$ that are not in $B^{\cap} C$
$A-\left(B^{\cap} C\right)=\{a, b, e, f\}$
(vi) (B - C)U(C - B) implies the union of sets B - C and C – B
$B-C=\{d, e\}$
$C-B=\{b, f\}$
$(B-C) \cup(C-B)=\{b, d, e, f\}$
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