1. Simplify the function using De Morgan's Theorem: (M + N)(M + N) - The problem is to simplify the expression (M + N)(M + N). - Using the idempotent law in Boolean algebra, we know that $X \cdot X = X$. - Therefore, $(M + N)(M + N) = M + N$. 2. Simplify the expression $x = (\overline{A} + B)(A + B + D)\overline{D}$ - The problem is to simplify $x = (\overline{A} + B)(A + B + D)\overline{D}$. - First, distribute terms: $$x = (\overline{A} + B)(A + B + D)\overline{D}$$ - Since $\overline{D}$ is ANDed, terms with $D$ will be zero: $$x = (\overline{A} + B)(A + B)\overline{D}$$ - Expand $(\overline{A} + B)(A + B)$: $$= \overline{A}A + \overline{A}B + BA + BB = 0 + \overline{A}B + AB + B$$ - Note $BB = B$ and $\overline{A}B + AB = B(\overline{A} + A) = B \cdot 1 = B$. - So the expression simplifies to: $$B + B = B$$ - Therefore: $$x = B \overline{D}$$ 3. Simplify $F(A,B,C) = A'BC + AB'C + ABC' + A'B'C$ using a 3-variable K-map - The problem is to simplify $F = A'BC + AB'C + ABC' + A'B'C$. - Write minterms: - $A'BC$ corresponds to $m_3$ - $AB'C$ corresponds to $m_5$ - $ABC'$ corresponds to $m_6$ - $A'B'C$ corresponds to $m_1$ - Plotting these on a 3-variable K-map and grouping: - Group 1: $m_1$ and $m_3$ share $B=1, C=1$ with $A$ varying. - Group 2: $m_3$ and $m_7$ (not in function) no, so group $m_3$ and $m_5$ share $A=1, C=1$ no. - Group 3: $m_5$ and $m_6$ share $A=1, B=0/1, C=1/0$ no. - Grouping $m_1$ and $m_3$ gives $B C$. - Grouping $m_5$ and $m_6$ gives $A B'$ and $A B C'$ separately. - Alternatively, sum of products can be simplified to: $$F = B C + A C + A B'$$ 4. Simplify $F(A,B,C,D) = \Sigma m(0,2,4,6,8)$ with don't cares $d = \Sigma m(10,11,12,13,14,15)$ using K-map - The problem is to simplify $F$ with minterms 0,2,4,6,8 and don't cares 10,11,12,13,14,15. - Plot these on a 4-variable K-map. - Group minterms and don't cares to form largest groups: - Group 1: minterms 8 and don't cares 10,11,12,13,14,15 form a large group covering $A=1$. - Group 2: minterms 0,2,4,6 form a group covering $A=0$ and $C=0$. - Simplified expression: $$F = A + B' C'$$ Summary: - (M + N)(M + N) = $M + N$ - $x = (\overline{A} + B)(A + B + D)\overline{D} = B \overline{D}$ - $F(A,B,C) = A'BC + AB'C + ABC' + A'B'C = B C + A C + A B'$ - $F(A,B,C,D) = \Sigma m(0,2,4,6,8)$ with don't cares $d = \Sigma m(10,11,12,13,14,15)$ simplifies to $F = A + B' C'$