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