De Morgan's Law Examples - Word Problems

# De Morgan's Law Examples

The De Morgan's laws state that complement of intersection of two sets is the union of their complements and complement of union of two sets id intersection of their complements.

$(A\cap B)' = A'\cup B'$
and
$(A\cup B)' = A'\cap B'$
Example 1:

If two sets $A = \{1, 2, 4, 5, 6\}$ and $B = \{2, 3, 4, 8\}$ then prove that $(A\cap B)' = A'\cup B'$.

Solution:

$(A\cap B)'$ = $\{2, 4\}' = \{1, 5, 6, 3, 8\}$

$A'\cup B'$ = $\{1, 2, 4, 5, 6\}'$ $\cup$ $\{2, 3, 4, 8\}'$

= $\{3, 8\}$ $\cup$ $\{1, 5, 6\}$

= $\{1, 5, 6, 3, 8\}$

Hence, it is proved that $(A\cap B)' = \{1, 5, 6, 3, 8\} = A'\cup B'$.

## De Morgan's Law Word Problems

Problem 1:

Let two sets be $P = \{a, e, d, j, k\}$ and $Q = \{k, l, m, e, n\}$ and $U = \{a, c, d, e, j, k, l, m, n\}$ then prove the the De Morgan's law of union using $P$ and $Q$.

Solution:

De Morgan's law of union states that,

$(P\cup Q)' = P'\cap Q'$, where P and Q are two sets.

$(P\cup Q)'$ = $(\{a, e, d, j, k\}$ $\cup$ $\{k, l, m, e, n\})'$ = $(\{a, e, d, j, k, l, m , n\})'$ = $\{c\}$

Now,

$P' = \{a, e, d, j, k\}' = \{c, l, m, n\}$

$Q' = \{k, l, m, e, n\}' = \{a, c, d, j\}$

$P'$ $\cap$ $Q' = \{c\}$

Hence, we can see that, $(P\cup Q)' = {c} = P'\cap Q'$
Problem 2:

Find the complement of $(A \cap B)\cup C$.

Solution:

Using De Morgan's law of union,

$((A \cap B)\cup C)' = (A\cap B)'\cap C'$

Again using De Morgan's law of intersection,

$(A\cap B)'\cap C' = A'\cup B'\cap C'$
Problem 3:

Prove that the complement of $(X\cup Y)'\cap Z$ equals $X'\cap (Y\cap Z)'$.

Solution:

Applying De Morgan's law of intersection we get,

$((X\cup Y)'\cap Z)' = (X\cup Y)'\cup Z'$

Now applying De Morgan's law of union on $(X\cup Y)'$ we get,

$(X\cup Y)'\cup Z' = X'\cap Y'\cup Z'$

Now,

$X'\cap Y'\cup Z' = X'\cap (Y'\cup Z')$

Using De Morgan's law of union,

$X'\cap (Y'\cup Z') = X'\cap (Y\cap Z)'$

Hence, it is proved that $((X\cup Y)'\cap Z)' = X'\cap (Y\cap Z)'$
Problem 4:

From the given Venn diagram, prove the De Morgan's law of union.

Solution:

De Morgan's law of union over the sets A and B states that,

$(A\cup B)' = A'\cap B'$

$(A\cup B)'$ = $\{x, y, z, a, b, p, q, r\}'$ = NULL

$A'\cap B'$ = $\{p, q, r\}$ $\cap$ $\{x, y, z\}$ = NULL

Hence, it is proved that $(A\cup B)' = A'\cap B'$.
Problem 5:

If universal set $U$ = $\{1, 2, 3, 4, 5, 6\}$, $A\ =\ \{1, 3\}$ and $B\ =\ \{4, 5, 6\}$ then prove De morgan's law of intersection.

Solution:

De morgan's law of intersection states that,

$(A\cap B)' = A'\cup B'$

$(A\cap B)'$ = $(\{1, 3\}$ $\cap$ $\{4, 5, 6\})'$ = (NULL)' = $\{1, 2, 3, 4, 5, 6\}$ (As complement of empty set is the universal set)

$A'\cup B'$ = $(\{1, 3\})$' $\cup$ $(\{4, 5, 6\})$'

= $\{2, 4, 5, 6\}$ $\cup$ $\{1, 2, 3\}$ = $\{1, 2, 3, 4, 5, 6\}$

Hence, it is proved that $(A\cap B)'\ =\ \{1, 2, 3, 4, 5, 6\}\ =\ A'\cup B'$